task_id
stringlengths 11
13
| test
stringlengths 117
1.8k
| entry_point
stringlengths 1
30
| gold_generation
stringlengths 16
864
| context
stringlengths 0
1.32k
| function_name
stringlengths 9
773
|
|---|---|---|---|---|---|
humaneval_100 | def check(candidate):
# Check some simple cases
assert candidate(3) == [3, 5, 7], "Test 3"
assert candidate(4) == [4,6,8,10], "Test 4"
assert candidate(5) == [5, 7, 9, 11, 13]
assert candidate(6) == [6, 8, 10, 12, 14, 16]
assert candidate(8) == [8, 10, 12, 14, 16, 18, 20, 22]
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| make_a_pile | return [n + 2*i for i in range(n)]
| Given a positive integer n, you have to make a pile of n levels of stones.
The first level has n stones.
The number of stones in the next level is:
- the next odd number if n is odd.
- the next even number if n is even.
Return the number of stones in each level in a list, where element at index
i represents the number of stones in the level (i+1).
Examples:
>>> make_a_pile(3)
[3, 5, 7] | def make_a_pile(n): |
humaneval_101 | def check(candidate):
# Check some simple cases
assert True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate("Hi, my name is John") == ["Hi", "my", "name", "is", "John"]
assert candidate("One, two, three, four, five, six") == ["One", "two", "three", "four", "five", "six"]
assert candidate("Hi, my name") == ["Hi", "my", "name"]
assert candidate("One,, two, three, four, five, six,") == ["One", "two", "three", "four", "five", "six"]
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate("") == []
assert candidate("ahmed , gamal") == ["ahmed", "gamal"]
| words_string | if not s:
return []
s_list = []
for letter in s:
if letter == ',':
s_list.append(' ')
else:
s_list.append(letter)
s_list = "".join(s_list)
return s_list.split()
| You will be given a string of words separated by commas or spaces. Your task is
to split the string into words and return an array of the words.
For example:
words_string("Hi, my name is John") == ["Hi", "my", "name", "is", "John"]
words_string("One, two, three, four, five, six") == ["One", "two", "three", "four", "five", "six"] | def words_string(s): |
humaneval_102 | def check(candidate):
# Check some simple cases
assert candidate(12, 15) == 14
assert candidate(13, 12) == -1
assert candidate(33, 12354) == 12354
assert candidate(5234, 5233) == -1
assert candidate(6, 29) == 28
assert candidate(27, 10) == -1
# Check some edge cases that are easy to work out by hand.
assert candidate(7, 7) == -1
assert candidate(546, 546) == 546
| choose_num | if x > y:
return -1
if y % 2 == 0:
return y
if x == y:
return -1
return y - 1
| This function takes two positive numbers x and y and returns the
biggest even integer number that is in the range [x, y] inclusive. If
there's no such number, then the function should return -1.
For example:
choose_num(12, 15) = 14
choose_num(13, 12) = -1 | def choose_num(x, y): |
humaneval_103 | def check(candidate):
# Check some simple cases
assert candidate(1, 5) == "0b11"
assert candidate(7, 13) == "0b1010"
assert candidate(964,977) == "0b1111001010"
assert candidate(996,997) == "0b1111100100"
assert candidate(560,851) == "0b1011000010"
assert candidate(185,546) == "0b101101110"
assert candidate(362,496) == "0b110101101"
assert candidate(350,902) == "0b1001110010"
assert candidate(197,233) == "0b11010111"
# Check some edge cases that are easy to work out by hand.
assert candidate(7, 5) == -1
assert candidate(5, 1) == -1
assert candidate(5, 5) == "0b101"
| rounded_avg | if m < n:
return -1
summation = 0
for i in range(n, m+1):
summation += i
return bin(round(summation/(m - n + 1)))
| You are given two positive integers n and m, and your task is to compute the
average of the integers from n through m (including n and m).
Round the answer to the nearest integer and convert that to binary.
If n is greater than m, return -1.
Example:
rounded_avg(1, 5) => "0b11"
rounded_avg(7, 5) => -1
rounded_avg(10, 20) => "0b1111"
rounded_avg(20, 33) => "0b11010" | def rounded_avg(n, m): |
humaneval_104 | def check(candidate):
# Check some simple cases
assert candidate([15, 33, 1422, 1]) == [1, 15, 33]
assert candidate([152, 323, 1422, 10]) == []
assert candidate([12345, 2033, 111, 151]) == [111, 151]
assert candidate([135, 103, 31]) == [31, 135]
# Check some edge cases that are easy to work out by hand.
assert True
| unique_digits | odd_digit_elements = []
for i in x:
if all (int(c) % 2 == 1 for c in str(i)):
odd_digit_elements.append(i)
return sorted(odd_digit_elements)
| Given a list of positive integers x. return a sorted list of all
elements that hasn't any even digit.
Note: Returned list should be sorted in increasing order.
For example:
>>> unique_digits([15, 33, 1422, 1])
[1, 15, 33]
>>> unique_digits([152, 323, 1422, 10])
[] | def unique_digits(x): |
humaneval_105 | def check(candidate):
# Check some simple cases
assert True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([2, 1, 1, 4, 5, 8, 2, 3]) == ["Eight", "Five", "Four", "Three", "Two", "Two", "One", "One"], "Error"
assert candidate([]) == [], "Error"
assert candidate([1, -1 , 55]) == ['One'], "Error"
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate([1, -1, 3, 2]) == ["Three", "Two", "One"]
assert candidate([9, 4, 8]) == ["Nine", "Eight", "Four"]
| by_length | dic = {
1: "One",
2: "Two",
3: "Three",
4: "Four",
5: "Five",
6: "Six",
7: "Seven",
8: "Eight",
9: "Nine",
}
sorted_arr = sorted(arr, reverse=True)
new_arr = []
for var in sorted_arr:
try:
new_arr.append(dic[var])
except:
pass
return new_arr
| Given an array of integers, sort the integers that are between 1 and 9 inclusive,
reverse the resulting array, and then replace each digit by its corresponding name from
"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine".
For example:
arr = [2, 1, 1, 4, 5, 8, 2, 3]
-> sort arr -> [1, 1, 2, 2, 3, 4, 5, 8]
-> reverse arr -> [8, 5, 4, 3, 2, 2, 1, 1]
return ["Eight", "Five", "Four", "Three", "Two", "Two", "One", "One"]
If the array is empty, return an empty array:
arr = []
return []
If the array has any strange number ignore it:
arr = [1, -1 , 55]
-> sort arr -> [-1, 1, 55]
-> reverse arr -> [55, 1, -1]
return = ['One'] | def by_length(arr): |
humaneval_106 | def check(candidate):
assert candidate(5) == [1, 2, 6, 24, 15]
assert candidate(7) == [1, 2, 6, 24, 15, 720, 28]
assert candidate(1) == [1]
assert candidate(3) == [1, 2, 6]
| f | ret = []
for i in range(1,n+1):
if i%2 == 0:
x = 1
for j in range(1,i+1): x *= j
ret += [x]
else:
x = 0
for j in range(1,i+1): x += j
ret += [x]
return ret
| Implement the function f that takes n as a parameter,
and returns a list of size n, such that the value of the element at index i is the factorial of i if i is even
or the sum of numbers from 1 to i otherwise.
i starts from 1.
the factorial of i is the multiplication of the numbers from 1 to i (1 * 2 * ... * i).
Example:
f(5) == [1, 2, 6, 24, 15] | def f(n): |
humaneval_107 | def check(candidate):
# Check some simple cases
assert candidate(123) == (8, 13)
assert candidate(12) == (4, 6)
assert candidate(3) == (1, 2)
assert candidate(63) == (6, 8)
assert candidate(25) == (5, 6)
assert candidate(19) == (4, 6)
assert candidate(9) == (4, 5), "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate(1) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"
| even_odd_palindrome | def is_palindrome(n):
return str(n) == str(n)[::-1]
even_palindrome_count = 0
odd_palindrome_count = 0
for i in range(1, n+1):
if i%2 == 1 and is_palindrome(i):
odd_palindrome_count += 1
elif i%2 == 0 and is_palindrome(i):
even_palindrome_count += 1
return (even_palindrome_count, odd_palindrome_count)
| Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Input: 3
Output: (1, 2)
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Input: 12
Output: (4, 6)
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively. | def even_odd_palindrome(n): |
humaneval_108 | def check(candidate):
# Check some simple cases
assert candidate([]) == 0
assert candidate([-1, -2, 0]) == 0
assert candidate([1, 1, 2, -2, 3, 4, 5]) == 6
assert candidate([1, 6, 9, -6, 0, 1, 5]) == 5
assert candidate([1, 100, 98, -7, 1, -1]) == 4
assert candidate([12, 23, 34, -45, -56, 0]) == 5
assert candidate([-0, 1**0]) == 1
assert candidate([1]) == 1
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| count_nums | def digits_sum(n):
neg = 1
if n < 0: n, neg = -1 * n, -1
n = [int(i) for i in str(n)]
n[0] = n[0] * neg
return sum(n)
return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))
| Write a function count_nums which takes an array of integers and returns
the number of elements which has a sum of digits > 0.
If a number is negative, then its first signed digit will be negative:
e.g. -123 has signed digits -1, 2, and 3.
>>> count_nums([]) == 0
>>> count_nums([-1, 11, -11]) == 1
>>> count_nums([1, 1, 2]) == 3 | def count_nums(arr): |
humaneval_109 | def check(candidate):
# Check some simple cases
assert candidate([3, 4, 5, 1, 2])==True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([3, 5, 10, 1, 2])==True
assert candidate([4, 3, 1, 2])==False
# Check some edge cases that are easy to work out by hand.
assert candidate([3, 5, 4, 1, 2])==False, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate([])==True
| move_one_ball | if len(arr)==0:
return True
sorted_array=sorted(arr)
my_arr=[]
min_value=min(arr)
min_index=arr.index(min_value)
my_arr=arr[min_index:]+arr[0:min_index]
for i in range(len(arr)):
if my_arr[i]!=sorted_array[i]:
return False
return True
| We have an array 'arr' of N integers arr[1], arr[2], ..., arr[N].The
numbers in the array will be randomly ordered. Your task is to determine if
it is possible to get an array sorted in non-decreasing order by performing
the following operation on the given array:
You are allowed to perform right shift operation any number of times.
One right shift operation means shifting all elements of the array by one
position in the right direction. The last element of the array will be moved to
the starting position in the array i.e. 0th index.
If it is possible to obtain the sorted array by performing the above operation
then return True else return False.
If the given array is empty then return True.
Note: The given list is guaranteed to have unique elements.
For Example:
move_one_ball([3, 4, 5, 1, 2])==>True
Explanation: By performin 2 right shift operations, non-decreasing order can
be achieved for the given array.
move_one_ball([3, 5, 4, 1, 2])==>False
Explanation:It is not possible to get non-decreasing order for the given
array by performing any number of right shift operations. | def move_one_ball(arr): |
humaneval_110 | def check(candidate):
# Check some simple cases
assert candidate([1, 2, 3, 4], [1, 2, 3, 4]) == "YES"
assert candidate([1, 2, 3, 4], [1, 5, 3, 4]) == "NO"
assert candidate([1, 2, 3, 4], [2, 1, 4, 3]) == "YES"
assert candidate([5, 7, 3], [2, 6, 4]) == "YES"
assert candidate([5, 7, 3], [2, 6, 3]) == "NO"
assert candidate([3, 2, 6, 1, 8, 9], [3, 5, 5, 1, 1, 1]) == "NO"
# Check some edge cases that are easy to work out by hand.
assert candidate([100, 200], [200, 200]) == "YES"
| exchange | odd = 0
even = 0
for i in lst1:
if i%2 == 1:
odd += 1
for i in lst2:
if i%2 == 0:
even += 1
if even >= odd:
return "YES"
return "NO"
| In this problem, you will implement a function that takes two lists of numbers,
and determines whether it is possible to perform an exchange of elements
between them to make lst1 a list of only even numbers.
There is no limit on the number of exchanged elements between lst1 and lst2.
If it is possible to exchange elements between the lst1 and lst2 to make
all the elements of lst1 to be even, return "YES".
Otherwise, return "NO".
For example:
exchange([1, 2, 3, 4], [1, 2, 3, 4]) => "YES"
exchange([1, 2, 3, 4], [1, 5, 3, 4]) => "NO"
It is assumed that the input lists will be non-empty. | def exchange(lst1, lst2): |
humaneval_111 | def check(candidate):
# Check some simple cases
assert candidate('a b b a') == {'a':2,'b': 2}, "This prints if this assert fails 1 (good for debugging!)"
assert candidate('a b c a b') == {'a': 2, 'b': 2}, "This prints if this assert fails 2 (good for debugging!)"
assert candidate('a b c d g') == {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'g': 1}, "This prints if this assert fails 3 (good for debugging!)"
assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, "This prints if this assert fails 4 (good for debugging!)"
assert candidate('b b b b a') == {'b': 4}, "This prints if this assert fails 5 (good for debugging!)"
assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, "This prints if this assert fails 6 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate('') == {}, "This prints if this assert fails 7 (also good for debugging!)"
assert candidate('a') == {'a': 1}, "This prints if this assert fails 8 (also good for debugging!)"
| histogram | dict1={}
list1=test.split(" ")
t=0
for i in list1:
if(list1.count(i)>t) and i!='':
t=list1.count(i)
if t>0:
for i in list1:
if(list1.count(i)==t):
dict1[i]=t
return dict1
| Given a string representing a space separated lowercase letters, return a dictionary
of the letter with the most repetition and containing the corresponding count.
If several letters have the same occurrence, return all of them.
Example:
histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
histogram('a b b a') == {'a': 2, 'b': 2}
histogram('a b c a b') == {'a': 2, 'b': 2}
histogram('b b b b a') == {'b': 4}
histogram('') == {} | def histogram(test): |
humaneval_112 | def check(candidate):
assert candidate("abcde","ae") == ('bcd',False)
assert candidate("abcdef", "b") == ('acdef',False)
assert candidate("abcdedcba","ab") == ('cdedc',True)
assert candidate("dwik","w") == ('dik',False)
assert candidate("a","a") == ('',True)
assert candidate("abcdedcba","") == ('abcdedcba',True)
assert candidate("abcdedcba","v") == ('abcdedcba',True)
assert candidate("vabba","v") == ('abba',True)
assert candidate("mamma", "mia") == ("", True)
| reverse_delete | s = ''.join([char for char in s if char not in c])
return (s,s[::-1] == s)
| Task
We are given two strings s and c, you have to deleted all the characters in s that are equal to any character in c
then check if the result string is palindrome.
A string is called palindrome if it reads the same backward as forward.
You should return a tuple containing the result string and True/False for the check.
Example
For s = "abcde", c = "ae", the result should be ('bcd',False)
For s = "abcdef", c = "b" the result should be ('acdef',False)
For s = "abcdedcba", c = "ab", the result should be ('cdedc',True) | def reverse_delete(s,c): |
humaneval_113 | def check(candidate):
# Check some simple cases
assert candidate(['1234567']) == ["the number of odd elements 4n the str4ng 4 of the 4nput."], "Test 1"
assert candidate(['3',"11111111"]) == ["the number of odd elements 1n the str1ng 1 of the 1nput.", "the number of odd elements 8n the str8ng 8 of the 8nput."], "Test 2"
assert candidate(['271', '137', '314']) == [
'the number of odd elements 2n the str2ng 2 of the 2nput.',
'the number of odd elements 3n the str3ng 3 of the 3nput.',
'the number of odd elements 2n the str2ng 2 of the 2nput.'
]
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| odd_count | res = []
for arr in lst:
n = sum(int(d)%2==1 for d in arr)
res.append("the number of odd elements " + str(n) + "n the str"+ str(n) +"ng "+ str(n) +" of the "+ str(n) +"nput.")
return res
| Given a list of strings, where each string consists of only digits, return a list.
Each element i of the output should be "the number of odd elements in the
string i of the input." where all the i's should be replaced by the number
of odd digits in the i'th string of the input.
>>> odd_count(['1234567'])
["the number of odd elements 4n the str4ng 4 of the 4nput."]
>>> odd_count(['3',"11111111"])
["the number of odd elements 1n the str1ng 1 of the 1nput.",
"the number of odd elements 8n the str8ng 8 of the 8nput."] | def odd_count(lst): |
humaneval_114 | def check(candidate):
# Check some simple cases
assert candidate([2, 3, 4, 1, 2, 4]) == 1, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([-1, -2, -3]) == -6
assert candidate([-1, -2, -3, 2, -10]) == -14
assert candidate([-9999999999999999]) == -9999999999999999
assert candidate([0, 10, 20, 1000000]) == 0
assert candidate([-1, -2, -3, 10, -5]) == -6
assert candidate([100, -1, -2, -3, 10, -5]) == -6
assert candidate([10, 11, 13, 8, 3, 4]) == 3
assert candidate([100, -33, 32, -1, 0, -2]) == -33
# Check some edge cases that are easy to work out by hand.
assert candidate([-10]) == -10, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate([7]) == 7
assert candidate([1, -1]) == -1
| minSubArraySum | max_sum = 0
s = 0
for num in nums:
s += -num
if (s < 0):
s = 0
max_sum = max(s, max_sum)
if max_sum == 0:
max_sum = max(-i for i in nums)
min_sum = -max_sum
return min_sum
| Given an array of integers nums, find the minimum sum of any non-empty sub-array
of nums.
Example
minSubArraySum([2, 3, 4, 1, 2, 4]) == 1
minSubArraySum([-1, -2, -3]) == -6 | def minSubArraySum(nums): |
humaneval_115 | def check(candidate):
# Check some simple cases
assert True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([[0,0,1,0], [0,1,0,0], [1,1,1,1]], 1) == 6, "Error"
assert candidate([[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]], 2) == 5, "Error"
assert candidate([[0,0,0], [0,0,0]], 5) == 0, "Error"
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate([[1,1,1,1], [1,1,1,1]], 2) == 4, "Error"
assert candidate([[1,1,1,1], [1,1,1,1]], 9) == 2, "Error"
| max_fill | return sum([math.ceil(sum(arr)/capacity) for arr in grid])
| You are given a rectangular grid of wells. Each row represents a single well,
and each 1 in a row represents a single unit of water.
Each well has a corresponding bucket that can be used to extract water from it,
and all buckets have the same capacity.
Your task is to use the buckets to empty the wells.
Output the number of times you need to lower the buckets.
Example 1:
Input:
grid : [[0,0,1,0], [0,1,0,0], [1,1,1,1]]
bucket_capacity : 1
Output: 6
Example 2:
Input:
grid : [[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]]
bucket_capacity : 2
Output: 5
Example 3:
Input:
grid : [[0,0,0], [0,0,0]]
bucket_capacity : 5
Output: 0
Constraints:
* all wells have the same length
* 1 <= grid.length <= 10^2
* 1 <= grid[:,1].length <= 10^2
* grid[i][j] -> 0 | 1
* 1 <= capacity <= 10 | def max_fill(grid, capacity):
import math |
humaneval_116 | def check(candidate):
# Check some simple cases
assert True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1,5,2,3,4]) == [1, 2, 4, 3, 5]
assert candidate([-2,-3,-4,-5,-6]) == [-4, -2, -6, -5, -3]
assert candidate([1,0,2,3,4]) == [0, 1, 2, 4, 3]
assert candidate([]) == []
assert candidate([2,5,77,4,5,3,5,7,2,3,4]) == [2, 2, 4, 4, 3, 3, 5, 5, 5, 7, 77]
assert candidate([3,6,44,12,32,5]) == [32, 3, 5, 6, 12, 44]
assert candidate([2,4,8,16,32]) == [2, 4, 8, 16, 32]
assert candidate([2,4,8,16,32]) == [2, 4, 8, 16, 32]
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| sort_array | return sorted(sorted(arr), key=lambda x: bin(x)[2:].count('1'))
| In this Kata, you have to sort an array of non-negative integers according to
number of ones in their binary representation in ascending order.
For similar number of ones, sort based on decimal value.
It must be implemented like this:
>>> sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]
>>> sort_array([-2, -3, -4, -5, -6]) == [-6, -5, -4, -3, -2]
>>> sort_array([1, 0, 2, 3, 4]) [0, 1, 2, 3, 4] | def sort_array(arr): |
humaneval_117 | def check(candidate):
# Check some simple cases
assert candidate("Mary had a little lamb", 4) == ["little"], "First test error: " + str(candidate("Mary had a little lamb", 4))
assert candidate("Mary had a little lamb", 3) == ["Mary", "lamb"], "Second test error: " + str(candidate("Mary had a little lamb", 3))
assert candidate("simple white space", 2) == [], "Third test error: " + str(candidate("simple white space", 2))
assert candidate("Hello world", 4) == ["world"], "Fourth test error: " + str(candidate("Hello world", 4))
assert candidate("Uncle sam", 3) == ["Uncle"], "Fifth test error: " + str(candidate("Uncle sam", 3))
# Check some edge cases that are easy to work out by hand.
assert candidate("", 4) == [], "1st edge test error: " + str(candidate("", 4))
assert candidate("a b c d e f", 1) == ["b", "c", "d", "f"], "2nd edge test error: " + str(candidate("a b c d e f", 1))
| select_words | result = []
for word in s.split():
n_consonants = 0
for i in range(0, len(word)):
if word[i].lower() not in ["a","e","i","o","u"]:
n_consonants += 1
if n_consonants == n:
result.append(word)
return result
| Given a string s and a natural number n, you have been tasked to implement
a function that returns a list of all words from string s that contain exactly
n consonants, in order these words appear in the string s.
If the string s is empty then the function should return an empty list.
Note: you may assume the input string contains only letters and spaces.
Examples:
select_words("Mary had a little lamb", 4) ==> ["little"]
select_words("Mary had a little lamb", 3) ==> ["Mary", "lamb"]
select_words("simple white space", 2) ==> []
select_words("Hello world", 4) ==> ["world"]
select_words("Uncle sam", 3) ==> ["Uncle"] | def select_words(s, n): |
humaneval_118 | def check(candidate):
# Check some simple cases
assert candidate("yogurt") == "u"
assert candidate("full") == "u"
assert candidate("easy") == ""
assert candidate("eAsy") == ""
assert candidate("ali") == ""
assert candidate("bad") == "a"
assert candidate("most") == "o"
assert candidate("ab") == ""
assert candidate("ba") == ""
assert candidate("quick") == ""
assert candidate("anime") == "i"
assert candidate("Asia") == ""
assert candidate("Above") == "o"
# Check some edge cases that are easy to work out by hand.
assert True
| get_closest_vowel | if len(word) < 3:
return ""
vowels = {"a", "e", "i", "o", "u", "A", "E", 'O', 'U', 'I'}
for i in range(len(word)-2, 0, -1):
if word[i] in vowels:
if (word[i+1] not in vowels) and (word[i-1] not in vowels):
return word[i]
return ""
| You are given a word. Your task is to find the closest vowel that stands between
two consonants from the right side of the word (case sensitive).
Vowels in the beginning and ending doesn't count. Return empty string if you didn't
find any vowel met the above condition.
You may assume that the given string contains English letter only.
Example:
get_closest_vowel("yogurt") ==> "u"
get_closest_vowel("FULL") ==> "U"
get_closest_vowel("quick") ==> ""
get_closest_vowel("ab") ==> "" | def get_closest_vowel(word): |
humaneval_119 | def check(candidate):
# Check some simple cases
assert candidate(['()(', ')']) == 'Yes'
assert candidate([')', ')']) == 'No'
assert candidate(['(()(())', '())())']) == 'No'
assert candidate([')())', '(()()(']) == 'Yes'
assert candidate(['(())))', '(()())((']) == 'Yes'
assert candidate(['()', '())']) == 'No'
assert candidate(['(()(', '()))()']) == 'Yes'
assert candidate(['((((', '((())']) == 'No'
assert candidate([')(()', '(()(']) == 'No'
assert candidate([')(', ')(']) == 'No'
# Check some edge cases that are easy to work out by hand.
assert candidate(['(', ')']) == 'Yes'
assert candidate([')', '(']) == 'Yes'
| match_parens | def check(s):
val = 0
for i in s:
if i == '(':
val = val + 1
else:
val = val - 1
if val < 0:
return False
return True if val == 0 else False
S1 = lst[0] + lst[1]
S2 = lst[1] + lst[0]
return 'Yes' if check(S1) or check(S2) else 'No'
| def match_parens(lst):
'''
You are given a list of two strings, both strings consist of open
parentheses '(' or close parentheses ')' only.
Your job is to check if it is possible to concatenate the two strings in
some order, that the resulting string will be good.
A string S is considered to be good if and only if all parentheses in S
are balanced. For example: the string '(())()' is good, while the string
'())' is not.
Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
Examples:
match_parens(['()(', ')']) == 'Yes'
match_parens([')', ')']) == 'No'
''' |
|
humaneval_120 | def check(candidate):
# Check some simple cases
assert candidate([-3, -4, 5], 3) == [-4, -3, 5]
assert candidate([4, -4, 4], 2) == [4, 4]
assert candidate([-3, 2, 1, 2, -1, -2, 1], 1) == [2]
assert candidate([123, -123, 20, 0 , 1, 2, -3], 3) == [2, 20, 123]
assert candidate([-123, 20, 0 , 1, 2, -3], 4) == [0, 1, 2, 20]
assert candidate([5, 15, 0, 3, -13, -8, 0], 7) == [-13, -8, 0, 0, 3, 5, 15]
assert candidate([-1, 0, 2, 5, 3, -10], 2) == [3, 5]
assert candidate([1, 0, 5, -7], 1) == [5]
assert candidate([4, -4], 2) == [-4, 4]
assert candidate([-10, 10], 2) == [-10, 10]
# Check some edge cases that are easy to work out by hand.
assert candidate([1, 2, 3, -23, 243, -400, 0], 0) == []
| maximum | if k == 0:
return []
arr.sort()
ans = arr[-k:]
return ans
| Given an array arr of integers and a positive integer k, return a sorted list
of length k with the maximum k numbers in arr.
Example 1:
Input: arr = [-3, -4, 5], k = 3
Output: [-4, -3, 5]
Example 2:
Input: arr = [4, -4, 4], k = 2
Output: [4, 4]
Example 3:
Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
Output: [2]
Note:
1. The length of the array will be in the range of [1, 1000].
2. The elements in the array will be in the range of [-1000, 1000].
3. 0 <= k <= len(arr) | def maximum(arr, k): |
humaneval_121 | def check(candidate):
# Check some simple cases
assert candidate([5, 8, 7, 1]) == 12
assert candidate([3, 3, 3, 3, 3]) == 9
assert candidate([30, 13, 24, 321]) == 0
assert candidate([5, 9]) == 5
assert candidate([2, 4, 8]) == 0
assert candidate([30, 13, 23, 32]) == 23
assert candidate([3, 13, 2, 9]) == 3
# Check some edge cases that are easy to work out by hand.
| solution | return sum([x for idx, x in enumerate(lst) if idx%2==0 and x%2==1])
| Given a non-empty list of integers, return the sum of all of the odd elements that are in even positions.
Examples
solution([5, 8, 7, 1]) ==> 12
solution([3, 3, 3, 3, 3]) ==> 9
solution([30, 13, 24, 321]) ==>0 | def solution(lst): |
humaneval_122 | def check(candidate):
# Check some simple cases
assert candidate([1,-2,-3,41,57,76,87,88,99], 3) == -4
assert candidate([111,121,3,4000,5,6], 2) == 0
assert candidate([11,21,3,90,5,6,7,8,9], 4) == 125
assert candidate([111,21,3,4000,5,6,7,8,9], 4) == 24, "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate([1], 1) == 1, "This prints if this assert fails 2 (also good for debugging!)"
| add_elements | return sum(elem for elem in arr[:k] if len(str(elem)) <= 2)
| Given a non-empty array of integers arr and an integer k, return
the sum of the elements with at most two digits from the first k elements of arr.
Example:
Input: arr = [111,21,3,4000,5,6,7,8,9], k = 4
Output: 24 # sum of 21 + 3
Constraints:
1. 1 <= len(arr) <= 100
2. 1 <= k <= len(arr) | def add_elements(arr, k): |
humaneval_123 | def check(candidate):
# Check some simple cases
assert candidate(14) == [1, 5, 7, 11, 13, 17]
assert candidate(5) == [1, 5]
assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"
| get_odd_collatz | if n%2==0:
odd_collatz = []
else:
odd_collatz = [n]
while n > 1:
if n % 2 == 0:
n = n/2
else:
n = n*3 + 1
if n%2 == 1:
odd_collatz.append(int(n))
return sorted(odd_collatz)
| Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.
The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
as follows: start with any positive integer n. Then each term is obtained from the
previous term as follows: if the previous term is even, the next term is one half of
the previous term. If the previous term is odd, the next term is 3 times the previous
term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.
Note:
1. Collatz(1) is [1].
2. returned list sorted in increasing order.
For example:
get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5. | def get_odd_collatz(n): |
humaneval_124 | def check(candidate):
# Check some simple cases
assert candidate('03-11-2000') == True
assert candidate('15-01-2012') == False
assert candidate('04-0-2040') == False
assert candidate('06-04-2020') == True
assert candidate('01-01-2007') == True
assert candidate('03-32-2011') == False
assert candidate('') == False
assert candidate('04-31-3000') == False
assert candidate('06-06-2005') == True
assert candidate('21-31-2000') == False
assert candidate('04-12-2003') == True
assert candidate('04122003') == False
assert candidate('20030412') == False
assert candidate('2003-04') == False
assert candidate('2003-04-12') == False
assert candidate('04-2003') == False
| valid_date | try:
date = date.strip()
month, day, year = date.split('-')
month, day, year = int(month), int(day), int(year)
if month < 1 or month > 12:
return False
if month in [1,3,5,7,8,10,12] and day < 1 or day > 31:
return False
if month in [4,6,9,11] and day < 1 or day > 30:
return False
if month == 2 and day < 1 or day > 29:
return False
except:
return False
return True
| You have to write a function which validates a given date string and
returns True if the date is valid otherwise False.
The date is valid if all of the following rules are satisfied:
1. The date string is not empty.
2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
3. The months should not be less than 1 or higher than 12.
4. The date should be in the format: mm-dd-yyyy
for example:
valid_date('03-11-2000') => True
valid_date('15-01-2012') => False
valid_date('04-0-2040') => False
valid_date('06-04-2020') => True
valid_date('06/04/2020') => False | def valid_date(date): |
humaneval_125 | def check(candidate):
assert candidate("Hello world!") == ["Hello","world!"]
assert candidate("Hello,world!") == ["Hello","world!"]
assert candidate("Hello world,!") == ["Hello","world,!"]
assert candidate("Hello,Hello,world !") == ["Hello,Hello,world","!"]
assert candidate("abcdef") == 3
assert candidate("aaabb") == 2
assert candidate("aaaBb") == 1
assert candidate("") == 0
| split_words | if " " in txt:
return txt.split()
elif "," in txt:
return txt.replace(',',' ').split()
else:
return len([i for i in txt if i.islower() and ord(i)%2 == 0])
| def split_words(txt):
'''
Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
Examples
split_words("Hello world!") ➞ ["Hello", "world!"]
split_words("Hello,world!") ➞ ["Hello", "world!"]
split_words("abcdef") == 3
''' |
|
humaneval_126 | def check(candidate):
# Check some simple cases
assert candidate([5]) == True
assert candidate([1, 2, 3, 4, 5]) == True
assert candidate([1, 3, 2, 4, 5]) == False
assert candidate([1, 2, 3, 4, 5, 6]) == True
assert candidate([1, 2, 3, 4, 5, 6, 7]) == True
assert candidate([1, 3, 2, 4, 5, 6, 7]) == False, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([]) == True, "This prints if this assert fails 2 (good for debugging!)"
assert candidate([1]) == True, "This prints if this assert fails 3 (good for debugging!)"
assert candidate([3, 2, 1]) == False, "This prints if this assert fails 4 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate([1, 2, 2, 2, 3, 4]) == False, "This prints if this assert fails 5 (good for debugging!)"
assert candidate([1, 2, 3, 3, 3, 4]) == False, "This prints if this assert fails 6 (good for debugging!)"
assert candidate([1, 2, 2, 3, 3, 4]) == True, "This prints if this assert fails 7 (good for debugging!)"
assert candidate([1, 2, 3, 4]) == True, "This prints if this assert fails 8 (good for debugging!)"
| is_sorted | count_digit = dict([(i, 0) for i in lst])
for i in lst:
count_digit[i]+=1
if any(count_digit[i] > 2 for i in lst):
return False
if all(lst[i-1] <= lst[i] for i in range(1, len(lst))):
return True
else:
return False
| def is_sorted(lst):
'''
Given a list of numbers, return whether or not they are sorted
in ascending order. If list has more than 1 duplicate of the same
number, return False. Assume no negative numbers and only integers.
Examples
is_sorted([5]) ➞ True
is_sorted([1, 2, 3, 4, 5]) ➞ True
is_sorted([1, 3, 2, 4, 5]) ➞ False
is_sorted([1, 2, 3, 4, 5, 6]) ➞ True
is_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True
is_sorted([1, 3, 2, 4, 5, 6, 7]) ➞ False
is_sorted([1, 2, 2, 3, 3, 4]) ➞ True
is_sorted([1, 2, 2, 2, 3, 4]) ➞ False
''' |
|
humaneval_127 | def check(candidate):
# Check some simple cases
assert candidate((1, 2), (2, 3)) == "NO"
assert candidate((-1, 1), (0, 4)) == "NO"
assert candidate((-3, -1), (-5, 5)) == "YES"
assert candidate((-2, 2), (-4, 0)) == "YES"
# Check some edge cases that are easy to work out by hand.
assert candidate((-11, 2), (-1, -1)) == "NO"
assert candidate((1, 2), (3, 5)) == "NO"
assert candidate((1, 2), (1, 2)) == "NO"
assert candidate((-2, -2), (-3, -2)) == "NO"
| intersection | def is_prime(num):
if num == 1 or num == 0:
return False
if num == 2:
return True
for i in range(2, num):
if num%i == 0:
return False
return True
l = max(interval1[0], interval2[0])
r = min(interval1[1], interval2[1])
length = r - l
if length > 0 and is_prime(length):
return "YES"
return "NO"
| You are given two intervals,
where each interval is a pair of integers. For example, interval = (start, end) = (1, 2).
The given intervals are closed which means that the interval (start, end)
includes both start and end.
For each given interval, it is assumed that its start is less or equal its end.
Your task is to determine whether the length of intersection of these two
intervals is a prime number.
Example, the intersection of the intervals (1, 3), (2, 4) is (2, 3)
which its length is 1, which not a prime number.
If the length of the intersection is a prime number, return "YES",
otherwise, return "NO".
If the two intervals don't intersect, return "NO".
[input/output] samples:
intersection((1, 2), (2, 3)) ==> "NO"
intersection((-1, 1), (0, 4)) ==> "NO"
intersection((-3, -1), (-5, 5)) ==> "YES" | def intersection(interval1, interval2): |
humaneval_128 | def check(candidate):
# Check some simple cases
assert True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1, 2, 2, -4]) == -9
assert candidate([0, 1]) == 0
assert candidate([1, 1, 1, 2, 3, -1, 1]) == -10
assert candidate([]) == None
assert candidate([2, 4,1, 2, -1, -1, 9]) == 20
assert candidate([-1, 1, -1, 1]) == 4
assert candidate([-1, 1, 1, 1]) == -4
assert candidate([-1, 1, 1, 0]) == 0
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| prod_signs | if not arr: return None
prod = 0 if 0 in arr else (-1) ** len(list(filter(lambda x: x < 0, arr)))
return prod * sum([abs(i) for i in arr])
| You are given an array arr of integers and you need to return
sum of magnitudes of integers multiplied by product of all signs
of each number in the array, represented by 1, -1 or 0.
Note: return None for empty arr.
Example:
>>> prod_signs([1, 2, 2, -4]) == -9
>>> prod_signs([0, 1]) == 0
>>> prod_signs([]) == None | def prod_signs(arr): |
humaneval_129 | def check(candidate):
# Check some simple cases
print
assert candidate([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3) == [1, 2, 1]
assert candidate([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1) == [1]
assert candidate([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]], 4) == [1, 2, 1, 2]
assert candidate([[6, 4, 13, 10], [5, 7, 12, 1], [3, 16, 11, 15], [8, 14, 9, 2]], 7) == [1, 10, 1, 10, 1, 10, 1]
assert candidate([[8, 14, 9, 2], [6, 4, 13, 15], [5, 7, 1, 12], [3, 10, 11, 16]], 5) == [1, 7, 1, 7, 1]
assert candidate([[11, 8, 7, 2], [5, 16, 14, 4], [9, 3, 15, 6], [12, 13, 10, 1]], 9) == [1, 6, 1, 6, 1, 6, 1, 6, 1]
assert candidate([[12, 13, 10, 1], [9, 3, 15, 6], [5, 16, 14, 4], [11, 8, 7, 2]], 12) == [1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6]
assert candidate([[2, 7, 4], [3, 1, 5], [6, 8, 9]], 8) == [1, 3, 1, 3, 1, 3, 1, 3]
assert candidate([[6, 1, 5], [3, 8, 9], [2, 7, 4]], 8) == [1, 5, 1, 5, 1, 5, 1, 5]
# Check some edge cases that are easy to work out by hand.
assert candidate([[1, 2], [3, 4]], 10) == [1, 2, 1, 2, 1, 2, 1, 2, 1, 2]
assert candidate([[1, 3], [3, 2]], 10) == [1, 3, 1, 3, 1, 3, 1, 3, 1, 3]
| minPath | n = len(grid)
val = n * n + 1
for i in range(n):
for j in range(n):
if grid[i][j] == 1:
temp = []
if i != 0:
temp.append(grid[i - 1][j])
if j != 0:
temp.append(grid[i][j - 1])
if i != n - 1:
temp.append(grid[i + 1][j])
if j != n - 1:
temp.append(grid[i][j + 1])
val = min(temp)
ans = []
for i in range(k):
if i % 2 == 0:
ans.append(1)
else:
ans.append(val)
return ans
| Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1] | def minPath(grid, k): |
humaneval_130 | def check(candidate):
# Check some simple cases
assert candidate(3) == [1, 3, 2.0, 8.0]
assert candidate(4) == [1, 3, 2.0, 8.0, 3.0]
assert candidate(5) == [1, 3, 2.0, 8.0, 3.0, 15.0]
assert candidate(6) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0]
assert candidate(7) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0]
assert candidate(8) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0, 5.0]
assert candidate(9) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0, 5.0, 35.0]
assert candidate(20) == [1, 3, 2.0, 8.0, 3.0, 15.0, 4.0, 24.0, 5.0, 35.0, 6.0, 48.0, 7.0, 63.0, 8.0, 80.0, 9.0, 99.0, 10.0, 120.0, 11.0]
# Check some edge cases that are easy to work out by hand.
assert candidate(0) == [1]
assert candidate(1) == [1, 3]
| tri | if n == 0:
return [1]
my_tri = [1, 3]
for i in range(2, n + 1):
if i % 2 == 0:
my_tri.append(i / 2 + 1)
else:
my_tri.append(my_tri[i - 1] + my_tri[i - 2] + (i + 3) / 2)
return my_tri
| Everyone knows Fibonacci sequence, it was studied deeply by mathematicians in
the last couple centuries. However, what people don't know is Tribonacci sequence.
Tribonacci sequence is defined by the recurrence:
tri(1) = 3
tri(n) = 1 + n / 2, if n is even.
tri(n) = tri(n - 1) + tri(n - 2) + tri(n + 1), if n is odd.
For example:
tri(2) = 1 + (2 / 2) = 2
tri(4) = 3
tri(3) = tri(2) + tri(1) + tri(4)
= 2 + 3 + 3 = 8
You are given a non-negative integer number n, you have to a return a list of the
first n + 1 numbers of the Tribonacci sequence.
Examples:
tri(3) = [1, 3, 2, 8] | def tri(n): |
humaneval_131 | def check(candidate):
# Check some simple cases
assert candidate(5) == 5
assert candidate(54) == 5
assert candidate(120) ==1
assert candidate(5014) == 5
assert candidate(98765) == 315
assert candidate(5576543) == 2625
# Check some edge cases that are easy to work out by hand.
assert candidate(2468) == 0
| digits | product = 1
odd_count = 0
for digit in str(n):
int_digit = int(digit)
if int_digit%2 == 1:
product= product*int_digit
odd_count+=1
if odd_count ==0:
return 0
else:
return product
| Given a positive integer n, return the product of the odd digits.
Return 0 if all digits are even.
For example:
digits(1) == 1
digits(4) == 0
digits(235) == 15 | def digits(n): |
humaneval_132 | def check(candidate):
# Check some simple cases
assert candidate('[[]]') == True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate('[]]]]]]][[[[[]') == False
assert candidate('[][]') == False
assert candidate(('[]')) == False
assert candidate('[[[[]]]]') == True
assert candidate('[]]]]]]]]]]') == False
assert candidate('[][][[]]') == True
assert candidate('[[]') == False
assert candidate('[]]') == False
assert candidate('[[]][[') == True
assert candidate('[[][]]') == True
# Check some edge cases that are easy to work out by hand.
assert candidate('') == False, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate('[[[[[[[[') == False
assert candidate(']]]]]]]]') == False
| is_nested | opening_bracket_index = []
closing_bracket_index = []
for i in range(len(string)):
if string[i] == '[':
opening_bracket_index.append(i)
else:
closing_bracket_index.append(i)
closing_bracket_index.reverse()
cnt = 0
i = 0
l = len(closing_bracket_index)
for idx in opening_bracket_index:
if i < l and idx < closing_bracket_index[i]:
cnt += 1
i += 1
return cnt >= 2
| def is_nested(string):
'''
Create a function that takes a string as input which contains only square brackets.
The function should return True if and only if there is a valid subsequence of brackets
where at least one bracket in the subsequence is nested.
is_nested('[[]]') ➞ True
is_nested('[]]]]]]][[[[[]') ➞ False
is_nested('[][]') ➞ False
is_nested('[]') ➞ False
is_nested('[[][]]') ➞ True
is_nested('[[]][[') ➞ True
''' |
|
humaneval_133 | def check(candidate):
# Check some simple cases
assert candidate([1,2,3])==14, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1.0,2,3])==14, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1,3,5,7])==84, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1.4,4.2,0])==29, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([-2.4,1,1])==6, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([100,1,15,2])==10230, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([10000,10000])==200000000, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([-1.4,4.6,6.3])==75, "This prints if this assert fails 1 (good for debugging!)"
assert candidate([-1.4,17.9,18.9,19.9])==1086, "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate([0])==0, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate([-1])==1, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate([-1,1,0])==2, "This prints if this assert fails 2 (also good for debugging!)"
| sum_squares | import math
squared = 0
for i in lst:
squared += math.ceil(i)**2
return squared
| You are given a list of numbers.
You need to return the sum of squared numbers in the given list,
round each element in the list to the upper int(Ceiling) first.
Examples:
For lst = [1,2,3] the output should be 14
For lst = [1,4,9] the output should be 98
For lst = [1,3,5,7] the output should be 84
For lst = [1.4,4.2,0] the output should be 29
For lst = [-2.4,1,1] the output should be 6 | def sum_squares(lst): |
humaneval_134 | def check(candidate):
# Check some simple cases
assert candidate("apple") == False
assert candidate("apple pi e") == True
assert candidate("eeeee") == False
assert candidate("A") == True
assert candidate("Pumpkin pie ") == False
assert candidate("Pumpkin pie 1") == False
assert candidate("") == False
assert candidate("eeeee e ") == False
assert candidate("apple pie") == False
assert candidate("apple pi e ") == False
# Check some edge cases that are easy to work out by hand.
assert True
| check_if_last_char_is_a_letter |
check = txt.split(' ')[-1]
return True if len(check) == 1 and (97 <= ord(check.lower()) <= 122) else False
| def check_if_last_char_is_a_letter(txt):
'''
Create a function that returns True if the last character
of a given string is an alphabetical character and is not
a part of a word, and False otherwise.
Note: "word" is a group of characters separated by space.
Examples:
check_if_last_char_is_a_letter("apple pie") ➞ False
check_if_last_char_is_a_letter("apple pi e") ➞ True
check_if_last_char_is_a_letter("apple pi e ") ➞ False
check_if_last_char_is_a_letter("") ➞ False
''' |
|
humaneval_135 | def check(candidate):
# Check some simple cases
assert candidate([1,2,4,3,5])==3
assert candidate([1,2,4,5])==-1
assert candidate([1,4,2,5,6,7,8,9,10])==2
assert candidate([4,8,5,7,3])==4
# Check some edge cases that are easy to work out by hand.
assert candidate([])==-1
| can_arrange | ind=-1
i=1
while i<len(arr):
if arr[i]<arr[i-1]:
ind=i
i+=1
return ind
| Create a function which returns the largest index of an element which
is not greater than or equal to the element immediately preceding it. If
no such element exists then return -1. The given array will not contain
duplicate values.
Examples:
can_arrange([1,2,4,3,5]) = 3
can_arrange([1,2,3]) = -1 | def can_arrange(arr): |
humaneval_136 | def check(candidate):
# Check some simple cases
assert candidate([2, 4, 1, 3, 5, 7]) == (None, 1)
assert candidate([2, 4, 1, 3, 5, 7, 0]) == (None, 1)
assert candidate([1, 3, 2, 4, 5, 6, -2]) == (-2, 1)
assert candidate([4, 5, 3, 6, 2, 7, -7]) == (-7, 2)
assert candidate([7, 3, 8, 4, 9, 2, 5, -9]) == (-9, 2)
assert candidate([]) == (None, None)
assert candidate([0]) == (None, None)
assert candidate([-1, -3, -5, -6]) == (-1, None)
assert candidate([-1, -3, -5, -6, 0]) == (-1, None)
assert candidate([-6, -4, -4, -3, 1]) == (-3, 1)
assert candidate([-6, -4, -4, -3, -100, 1]) == (-3, 1)
# Check some edge cases that are easy to work out by hand.
assert True
| largest_smallest_integers | smallest = list(filter(lambda x: x < 0, lst))
largest = list(filter(lambda x: x > 0, lst))
return (max(smallest) if smallest else None, min(largest) if largest else None)
| def largest_smallest_integers(lst):
'''
Create a function that returns a tuple (a, b), where 'a' is
the largest of negative integers, and 'b' is the smallest
of positive integers in a list.
If there is no negative or positive integers, return them as None.
Examples:
largest_smallest_integers([2, 4, 1, 3, 5, 7]) == (None, 1)
largest_smallest_integers([]) == (None, None)
largest_smallest_integers([0]) == (None, None)
''' |
|
humaneval_137 | def check(candidate):
# Check some simple cases
assert candidate(1, 2) == 2
assert candidate(1, 2.5) == 2.5
assert candidate(2, 3) == 3
assert candidate(5, 6) == 6
assert candidate(1, "2,3") == "2,3"
assert candidate("5,1", "6") == "6"
assert candidate("1", "2") == "2"
assert candidate("1", 1) == None
# Check some edge cases that are easy to work out by hand.
assert True
| compare_one | temp_a, temp_b = a, b
if isinstance(temp_a, str): temp_a = temp_a.replace(',','.')
if isinstance(temp_b, str): temp_b = temp_b.replace(',','.')
if float(temp_a) == float(temp_b): return None
return a if float(temp_a) > float(temp_b) else b
| Create a function that takes integers, floats, or strings representing
real numbers, and returns the larger variable in its given variable type.
Return None if the values are equal.
Note: If a real number is represented as a string, the floating point might be . or ,
compare_one(1, 2.5) ➞ 2.5
compare_one(1, "2,3") ➞ "2,3"
compare_one("5,1", "6") ➞ "6"
compare_one("1", 1) ➞ None | def compare_one(a, b): |
humaneval_138 | def check(candidate):
assert candidate(4) == False
assert candidate(6) == False
assert candidate(8) == True
assert candidate(10) == True
assert candidate(11) == False
assert candidate(12) == True
assert candidate(13) == False
assert candidate(16) == True
| is_equal_to_sum_even | return n%2 == 0 and n >= 8
| Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers
Example
is_equal_to_sum_even(4) == False
is_equal_to_sum_even(6) == False
is_equal_to_sum_even(8) == True | def is_equal_to_sum_even(n): |
humaneval_139 | def check(candidate):
# Check some simple cases
assert candidate(4) == 288, "Test 4"
assert candidate(5) == 34560, "Test 5"
assert candidate(7) == 125411328000, "Test 7"
# Check some edge cases that are easy to work out by hand.
assert candidate(1) == 1, "Test 1"
| special_factorial | fact_i = 1
special_fact = 1
for i in range(1, n+1):
fact_i *= i
special_fact *= fact_i
return special_fact
| The Brazilian factorial is defined as:
brazilian_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!
where n > 0
For example:
>>> special_factorial(4)
288
The function will receive an integer as input and should return the special
factorial of this integer. | def special_factorial(n): |
humaneval_140 | def check(candidate):
# Check some simple cases
assert candidate("Example") == "Example", "This prints if this assert fails 1 (good for debugging!)"
assert candidate("Mudasir Hanif ") == "Mudasir_Hanif_", "This prints if this assert fails 2 (good for debugging!)"
assert candidate("Yellow Yellow Dirty Fellow") == "Yellow_Yellow__Dirty__Fellow", "This prints if this assert fails 3 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate("Exa mple") == "Exa-mple", "This prints if this assert fails 4 (good for debugging!)"
assert candidate(" Exa 1 2 2 mple") == "-Exa_1_2_2_mple", "This prints if this assert fails 4 (good for debugging!)"
| fix_spaces | new_text = ""
i = 0
start, end = 0, 0
while i < len(text):
if text[i] == " ":
end += 1
else:
if end - start > 2:
new_text += "-"+text[i]
elif end - start > 0:
new_text += "_"*(end - start)+text[i]
else:
new_text += text[i]
start, end = i+1, i+1
i+=1
if end - start > 2:
new_text += "-"
elif end - start > 0:
new_text += "_"
return new_text
| Given a string text, replace all spaces in it with underscores,
and if a string has more than 2 consecutive spaces,
then replace all consecutive spaces with -
fix_spaces("Example") == "Example"
fix_spaces("Example 1") == "Example_1"
fix_spaces(" Example 2") == "_Example_2"
fix_spaces(" Example 3") == "_Example-3" | def fix_spaces(text): |
humaneval_141 | def check(candidate):
# Check some simple cases
assert candidate("example.txt") == 'Yes'
assert candidate("1example.dll") == 'No'
assert candidate('s1sdf3.asd') == 'No'
assert candidate('K.dll') == 'Yes'
assert candidate('MY16FILE3.exe') == 'Yes'
assert candidate('His12FILE94.exe') == 'No'
assert candidate('_Y.txt') == 'No'
assert candidate('?aREYA.exe') == 'No'
assert candidate('/this_is_valid.dll') == 'No'
assert candidate('this_is_valid.wow') == 'No'
assert candidate('this_is_valid.txt') == 'Yes'
assert candidate('this_is_valid.txtexe') == 'No'
assert candidate('#this2_i4s_5valid.ten') == 'No'
assert candidate('@this1_is6_valid.exe') == 'No'
assert candidate('this_is_12valid.6exe4.txt') == 'No'
assert candidate('all.exe.txt') == 'No'
assert candidate('I563_No.exe') == 'Yes'
assert candidate('Is3youfault.txt') == 'Yes'
assert candidate('no_one#knows.dll') == 'Yes'
assert candidate('1I563_Yes3.exe') == 'No'
assert candidate('I563_Yes3.txtt') == 'No'
assert candidate('final..txt') == 'No'
assert candidate('final132') == 'No'
assert candidate('_f4indsartal132.') == 'No'
# Check some edge cases that are easy to work out by hand.
assert candidate('.txt') == 'No'
assert candidate('s.') == 'No'
| file_name_check | suf = ['txt', 'exe', 'dll']
lst = file_name.split(sep='.')
if len(lst) != 2:
return 'No'
if not lst[1] in suf:
return 'No'
if len(lst[0]) == 0:
return 'No'
if not lst[0][0].isalpha():
return 'No'
t = len([x for x in lst[0] if x.isdigit()])
if t > 3:
return 'No'
return 'Yes'
| Create a function which takes a string representing a file's name, and returns
'Yes' if the the file's name is valid, and returns 'No' otherwise.
A file's name is considered to be valid if and only if all the following conditions
are met:
- There should not be more than three digits ('0'-'9') in the file's name.
- The file's name contains exactly one dot '.'
- The substring before the dot should not be empty, and it starts with a letter from
the latin alphapet ('a'-'z' and 'A'-'Z').
- The substring after the dot should be one of these: ['txt', 'exe', 'dll']
Examples:
file_name_check("example.txt") # => 'Yes'
file_name_check("1example.dll") # => 'No' (the name should start with a latin alphapet letter) | def file_name_check(file_name): |
humaneval_142 | def check(candidate):
# Check some simple cases
assert candidate([1,2,3]) == 6
assert candidate([1,4,9]) == 14
assert candidate([]) == 0
assert candidate([1,1,1,1,1,1,1,1,1]) == 9
assert candidate([-1,-1,-1,-1,-1,-1,-1,-1,-1]) == -3
assert candidate([0]) == 0
assert candidate([-1,-5,2,-1,-5]) == -126
assert candidate([-56,-99,1,0,-2]) == 3030
assert candidate([-1,0,0,0,0,0,0,0,-1]) == 0
assert candidate([-16, -9, -2, 36, 36, 26, -20, 25, -40, 20, -4, 12, -26, 35, 37]) == -14196
assert candidate([-1, -3, 17, -1, -15, 13, -1, 14, -14, -12, -5, 14, -14, 6, 13, 11, 16, 16, 4, 10]) == -1448
# Don't remove this line:
| sum_squares | result =[]
for i in range(len(lst)):
if i %3 == 0:
result.append(lst[i]**2)
elif i % 4 == 0 and i%3 != 0:
result.append(lst[i]**3)
else:
result.append(lst[i])
return sum(result)
| "
This function will take a list of integers. For all entries in the list, the function shall square the integer entry if its index is a
multiple of 3 and will cube the integer entry if its index is a multiple of 4 and not a multiple of 3. The function will not
change the entries in the list whose indexes are not a multiple of 3 or 4. The function shall then return the sum of all entries.
Examples:
For lst = [1,2,3] the output should be 6
For lst = [] the output should be 0
For lst = [-1,-5,2,-1,-5] the output should be -126 | def sum_squares(lst): |
humaneval_143 | def check(candidate):
# Check some simple cases
assert candidate("This is a test") == "is"
assert candidate("lets go for swimming") == "go for"
assert candidate("there is no place available here") == "there is no place"
assert candidate("Hi I am Hussein") == "Hi am Hussein"
assert candidate("go for it") == "go for it"
# Check some edge cases that are easy to work out by hand.
assert candidate("here") == ""
assert candidate("here is") == "is"
| words_in_sentence | new_lst = []
for word in sentence.split():
flg = 0
if len(word) == 1:
flg = 1
for i in range(2, len(word)):
if len(word)%i == 0:
flg = 1
if flg == 0 or len(word) == 2:
new_lst.append(word)
return " ".join(new_lst)
| You are given a string representing a sentence,
the sentence contains some words separated by a space,
and you have to return a string that contains the words from the original sentence,
whose lengths are prime numbers,
the order of the words in the new string should be the same as the original one.
Example 1:
Input: sentence = "This is a test"
Output: "is"
Example 2:
Input: sentence = "lets go for swimming"
Output: "go for"
Constraints:
* 1 <= len(sentence) <= 100
* sentence contains only letters | def words_in_sentence(sentence): |
humaneval_144 | def check(candidate):
# Check some simple cases
assert candidate("1/5", "5/1") == True, 'test1'
assert candidate("1/6", "2/1") == False, 'test2'
assert candidate("5/1", "3/1") == True, 'test3'
assert candidate("7/10", "10/2") == False, 'test4'
assert candidate("2/10", "50/10") == True, 'test5'
assert candidate("7/2", "4/2") == True, 'test6'
assert candidate("11/6", "6/1") == True, 'test7'
assert candidate("2/3", "5/2") == False, 'test8'
assert candidate("5/2", "3/5") == False, 'test9'
assert candidate("2/4", "8/4") == True, 'test10'
# Check some edge cases that are easy to work out by hand.
assert candidate("2/4", "4/2") == True, 'test11'
assert candidate("1/5", "5/1") == True, 'test12'
assert candidate("1/5", "1/5") == False, 'test13'
| simplify | a, b = x.split("/")
c, d = n.split("/")
numerator = int(a) * int(c)
denom = int(b) * int(d)
if (numerator/denom == int(numerator/denom)):
return True
return False
| Your task is to implement a function that will simplify the expression
x * n. The function returns True if x * n evaluates to a whole number and False
otherwise. Both x and n, are string representation of a fraction, and have the following format,
<numerator>/<denominator> where both numerator and denominator are positive whole numbers.
You can assume that x, and n are valid fractions, and do not have zero as denominator.
simplify("1/5", "5/1") = True
simplify("1/6", "2/1") = False
simplify("7/10", "10/2") = False | def simplify(x, n): |
humaneval_145 | def check(candidate):
# Check some simple cases
assert candidate([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]
assert candidate([1234,423,463,145,2,423,423,53,6,37,3457,3,56,0,46]) == [0, 2, 3, 6, 53, 423, 423, 423, 1234, 145, 37, 46, 56, 463, 3457]
assert candidate([]) == []
assert candidate([1, -11, -32, 43, 54, -98, 2, -3]) == [-3, -32, -98, -11, 1, 2, 43, 54]
assert candidate([1,2,3,4,5,6,7,8,9,10,11]) == [1, 10, 2, 11, 3, 4, 5, 6, 7, 8, 9]
assert candidate([0,6,6,-76,-21,23,4]) == [-76, -21, 0, 4, 23, 6, 6]
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| order_by_points | def digits_sum(n):
neg = 1
if n < 0: n, neg = -1 * n, -1
n = [int(i) for i in str(n)]
n[0] = n[0] * neg
return sum(n)
return sorted(nums, key=digits_sum)
| Write a function which sorts the given list of integers
in ascending order according to the sum of their digits.
Note: if there are several items with similar sum of their digits,
order them based on their index in original list.
For example:
>>> order_by_points([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]
>>> order_by_points([]) == [] | def order_by_points(nums): |
humaneval_146 | def check(candidate):
# Check some simple cases
assert candidate([5, -2, 1, -5]) == 0
assert candidate([15, -73, 14, -15]) == 1
assert candidate([33, -2, -3, 45, 21, 109]) == 2
assert candidate([43, -12, 93, 125, 121, 109]) == 4
assert candidate([71, -2, -33, 75, 21, 19]) == 3
# Check some edge cases that are easy to work out by hand.
assert candidate([1]) == 0
assert candidate([]) == 0
| specialFilter |
count = 0
for num in nums:
if num > 10:
odd_digits = (1, 3, 5, 7, 9)
number_as_string = str(num)
if int(number_as_string[0]) in odd_digits and int(number_as_string[-1]) in odd_digits:
count += 1
return count
| Write a function that takes an array of numbers as input and returns
the number of elements in the array that are greater than 10 and both
first and last digits of a number are odd (1, 3, 5, 7, 9).
For example:
specialFilter([15, -73, 14, -15]) => 1
specialFilter([33, -2, -3, 45, 21, 109]) => 2 | def specialFilter(nums): |
humaneval_147 | def check(candidate):
assert candidate(5) == 1
assert candidate(6) == 4
assert candidate(10) == 36
assert candidate(100) == 53361
| get_max_triples | A = [i*i - i + 1 for i in range(1,n+1)]
ans = []
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
if (A[i]+A[j]+A[k])%3 == 0:
ans += [(A[i],A[j],A[k])]
return len(ans)
| You are given a positive integer n. You have to create an integer array a of length n.
For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.
Return the number of triples (a[i], a[j], a[k]) of a where i < j < k,
and a[i] + a[j] + a[k] is a multiple of 3.
Example :
Input: n = 5
Output: 1
Explanation:
a = [1, 3, 7, 13, 21]
The only valid triple is (1, 7, 13). | def get_max_triples(n): |
humaneval_148 | def check(candidate):
# Check some simple cases
assert candidate("Jupiter", "Neptune") == ("Saturn", "Uranus"), "First test error: " + str(len(candidate("Jupiter", "Neptune")))
assert candidate("Earth", "Mercury") == ("Venus",), "Second test error: " + str(candidate("Earth", "Mercury"))
assert candidate("Mercury", "Uranus") == ("Venus", "Earth", "Mars", "Jupiter", "Saturn"), "Third test error: " + str(candidate("Mercury", "Uranus"))
assert candidate("Neptune", "Venus") == ("Earth", "Mars", "Jupiter", "Saturn", "Uranus"), "Fourth test error: " + str(candidate("Neptune", "Venus"))
# Check some edge cases that are easy to work out by hand.
assert candidate("Earth", "Earth") == ()
assert candidate("Mars", "Earth") == ()
assert candidate("Jupiter", "Makemake") == ()
| bf | planet_names = ("Mercury", "Venus", "Earth", "Mars", "Jupiter", "Saturn", "Uranus", "Neptune")
if planet1 not in planet_names or planet2 not in planet_names or planet1 == planet2:
return ()
planet1_index = planet_names.index(planet1)
planet2_index = planet_names.index(planet2)
if planet1_index < planet2_index:
return (planet_names[planet1_index + 1: planet2_index])
else:
return (planet_names[planet2_index + 1 : planet1_index])
| def bf(planet1, planet2):
'''
There are eight planets in our solar system: the closerst to the Sun
is Mercury, the next one is Venus, then Earth, Mars, Jupiter, Saturn,
Uranus, Neptune.
Write a function that takes two planet names as strings planet1 and planet2.
The function should return a tuple containing all planets whose orbits are
located between the orbit of planet1 and the orbit of planet2, sorted by
the proximity to the sun.
The function should return an empty tuple if planet1 or planet2
are not correct planet names.
Examples
bf("Jupiter", "Neptune") ==> ("Saturn", "Uranus")
bf("Earth", "Mercury") ==> ("Venus")
bf("Mercury", "Uranus") ==> ("Venus", "Earth", "Mars", "Jupiter", "Saturn")
''' |
|
humaneval_149 | def check(candidate):
# Check some simple cases
assert candidate(["aa", "a", "aaa"]) == ["aa"]
assert candidate(["school", "AI", "asdf", "b"]) == ["AI", "asdf", "school"]
assert candidate(["d", "b", "c", "a"]) == []
assert candidate(["d", "dcba", "abcd", "a"]) == ["abcd", "dcba"]
# Check some edge cases that are easy to work out by hand.
assert candidate(["AI", "ai", "au"]) == ["AI", "ai", "au"]
assert candidate(["a", "b", "b", "c", "c", "a"]) == []
assert candidate(['aaaa', 'bbbb', 'dd', 'cc']) == ["cc", "dd", "aaaa", "bbbb"]
| sorted_list_sum | lst.sort()
new_lst = []
for i in lst:
if len(i)%2 == 0:
new_lst.append(i)
return sorted(new_lst, key=len)
| Write a function that accepts a list of strings as a parameter,
deletes the strings that have odd lengths from it,
and returns the resulted list with a sorted order,
The list is always a list of strings and never an array of numbers,
and it may contain duplicates.
The order of the list should be ascending by length of each word, and you
should return the list sorted by that rule.
If two words have the same length, sort the list alphabetically.
The function should return a list of strings in sorted order.
You may assume that all words will have the same length.
For example:
assert list_sort(["aa", "a", "aaa"]) => ["aa"]
assert list_sort(["ab", "a", "aaa", "cd"]) => ["ab", "cd"] | def sorted_list_sum(lst): |
humaneval_150 | def check(candidate):
# Check some simple cases
assert candidate(7, 34, 12) == 34
assert candidate(15, 8, 5) == 5
assert candidate(3, 33, 5212) == 33
assert candidate(1259, 3, 52) == 3
assert candidate(7919, -1, 12) == -1
assert candidate(3609, 1245, 583) == 583
assert candidate(91, 56, 129) == 129
assert candidate(6, 34, 1234) == 1234
# Check some edge cases that are easy to work out by hand.
assert candidate(1, 2, 0) == 0
assert candidate(2, 2, 0) == 2
| x_or_y | if n == 1:
return y
for i in range(2, n):
if n % i == 0:
return y
break
else:
return x
| A simple program which should return the value of x if n is
a prime number and should return the value of y otherwise.
Examples:
for x_or_y(7, 34, 12) == 34
for x_or_y(15, 8, 5) == 5 | def x_or_y(n, x, y): |
humaneval_151 | def check(candidate):
# Check some simple cases
assert candidate([]) == 0 , "This prints if this assert fails 1 (good for debugging!)"
assert candidate([5, 4]) == 25 , "This prints if this assert fails 2 (good for debugging!)"
assert candidate([0.1, 0.2, 0.3]) == 0 , "This prints if this assert fails 3 (good for debugging!)"
assert candidate([-10, -20, -30]) == 0 , "This prints if this assert fails 4 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert candidate([-1, -2, 8]) == 0, "This prints if this assert fails 5 (also good for debugging!)"
assert candidate([0.2, 3, 5]) == 34, "This prints if this assert fails 6 (also good for debugging!)"
lst = list(range(-99, 100, 2))
odd_sum = sum([i**2 for i in lst if i%2!=0 and i > 0])
assert candidate(lst) == odd_sum , "This prints if this assert fails 7 (good for debugging!)"
| double_the_difference | return sum([i**2 for i in lst if i > 0 and i%2!=0 and "." not in str(i)])
| def double_the_difference(lst):
'''
Given a list of numbers, return the sum of squares of the numbers
in the list that are odd. Ignore numbers that are negative or not integers.
double_the_difference([1, 3, 2, 0]) == 1 + 9 + 0 + 0 = 10
double_the_difference([-1, -2, 0]) == 0
double_the_difference([9, -2]) == 81
double_the_difference([0]) == 0
If the input list is empty, return 0.
''' |
|
humaneval_152 | def check(candidate):
# Check some simple cases
assert candidate([1,2,3,4,5,1],[1,2,3,4,2,-2])==[0,0,0,0,3,3], "This prints if this assert fails 1 (good for debugging!)"
assert candidate([0,0,0,0,0,0],[0,0,0,0,0,0])==[0,0,0,0,0,0], "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1,2,3],[-1,-2,-3])==[2,4,6], "This prints if this assert fails 1 (good for debugging!)"
assert candidate([1,2,3,5],[-1,2,3,4])==[2,0,0,1], "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| compare | return [abs(x-y) for x,y in zip(game,guess)]
| I think we all remember that feeling when the result of some long-awaited
event is finally known. The feelings and thoughts you have at that moment are
definitely worth noting down and comparing.
Your task is to determine if a person correctly guessed the results of a number of matches.
You are given two arrays of scores and guesses of equal length, where each index shows a match.
Return an array of the same length denoting how far off each guess was. If they have guessed correctly,
the value is 0, and if not, the value is the absolute difference between the guess and the score.
example:
compare([1,2,3,4,5,1],[1,2,3,4,2,-2]) -> [0,0,0,0,3,3]
compare([0,5,0,0,0,4],[4,1,1,0,0,-2]) -> [4,4,1,0,0,6] | def compare(game,guess): |
humaneval_153 | def check(candidate):
# Check some simple cases
assert candidate('Watashi', ['tEN', 'niNE', 'eIGHt8OKe']) == 'Watashi.eIGHt8OKe'
assert candidate('Boku123', ['nani', 'NazeDa', 'YEs.WeCaNe', '32145tggg']) == 'Boku123.YEs.WeCaNe'
assert candidate('__YESIMHERE', ['t', 'eMptY', 'nothing', 'zeR00', 'NuLl__', '123NoooneB321']) == '__YESIMHERE.NuLl__'
assert candidate('K', ['Ta', 'TAR', 't234An', 'cosSo']) == 'K.TAR'
assert candidate('__HAHA', ['Tab', '123', '781345', '-_-']) == '__HAHA.123'
assert candidate('YameRore', ['HhAas', 'okIWILL123', 'WorkOut', 'Fails', '-_-']) == 'YameRore.okIWILL123'
assert candidate('finNNalLLly', ['Die', 'NowW', 'Wow', 'WoW']) == 'finNNalLLly.WoW'
# Check some edge cases that are easy to work out by hand.
assert candidate('_', ['Bb', '91245']) == '_.Bb'
assert candidate('Sp', ['671235', 'Bb']) == 'Sp.671235'
| Strongest_Extension | strong = extensions[0]
my_val = len([x for x in extensions[0] if x.isalpha() and x.isupper()]) - len([x for x in extensions[0] if x.isalpha() and x.islower()])
for s in extensions:
val = len([x for x in s if x.isalpha() and x.isupper()]) - len([x for x in s if x.isalpha() and x.islower()])
if val > my_val:
strong = s
my_val = val
ans = class_name + "." + strong
return ans
| You will be given the name of a class (a string) and a list of extensions.
The extensions are to be used to load additional classes to the class. The
strength of the extension is as follows: Let CAP be the number of the uppercase
letters in the extension's name, and let SM be the number of lowercase letters
in the extension's name, the strength is given by the fraction CAP - SM.
You should find the strongest extension and return a string in this
format: ClassName.StrongestExtensionName.
If there are two or more extensions with the same strength, you should
choose the one that comes first in the list.
For example, if you are given "Slices" as the class and a list of the
extensions: ['SErviNGSliCes', 'Cheese', 'StuFfed'] then you should
return 'Slices.SErviNGSliCes' since 'SErviNGSliCes' is the strongest extension
(its strength is -1).
Example:
for Strongest_Extension('my_class', ['AA', 'Be', 'CC']) == 'my_class.AA' | def Strongest_Extension(class_name, extensions): |
humaneval_154 | def check(candidate):
# Check some simple cases
#assert True, "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
#assert True, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate("xyzw","xyw") == False , "test #0"
assert candidate("yello","ell") == True , "test #1"
assert candidate("whattup","ptut") == False , "test #2"
assert candidate("efef","fee") == True , "test #3"
assert candidate("abab","aabb") == False , "test #4"
assert candidate("winemtt","tinem") == True , "test #5"
| cycpattern_check | l = len(b)
pat = b + b
for i in range(len(a) - l + 1):
for j in range(l + 1):
if a[i:i+l] == pat[j:j+l]:
return True
return False
| You are given 2 words. You need to return True if the second word or any of its rotations is a substring in the first word
cycpattern_check("abcd","abd") => False
cycpattern_check("hello","ell") => True
cycpattern_check("whassup","psus") => False
cycpattern_check("abab","baa") => True
cycpattern_check("efef","eeff") => False
cycpattern_check("himenss","simen") => True | def cycpattern_check(a , b): |
humaneval_155 | def check(candidate):
# Check some simple cases
assert candidate(7) == (0, 1)
assert candidate(-78) == (1, 1)
assert candidate(3452) == (2, 2)
assert candidate(346211) == (3, 3)
assert candidate(-345821) == (3, 3)
assert candidate(-2) == (1, 0)
assert candidate(-45347) == (2, 3)
assert candidate(0) == (1, 0)
# Check some edge cases that are easy to work out by hand.
assert True
| even_odd_count | even_count = 0
odd_count = 0
for i in str(abs(num)):
if int(i)%2==0:
even_count +=1
else:
odd_count +=1
return (even_count, odd_count)
| Given an integer. return a tuple that has the number of even and odd digits respectively.
Example:
even_odd_count(-12) ==> (1, 1)
even_odd_count(123) ==> (1, 2) | def even_odd_count(num): |
humaneval_156 | def check(candidate):
# Check some simple cases
assert candidate(19) == 'xix'
assert candidate(152) == 'clii'
assert candidate(251) == 'ccli'
assert candidate(426) == 'cdxxvi'
assert candidate(500) == 'd'
assert candidate(1) == 'i'
assert candidate(4) == 'iv'
assert candidate(43) == 'xliii'
assert candidate(90) == 'xc'
assert candidate(94) == 'xciv'
assert candidate(532) == 'dxxxii'
assert candidate(900) == 'cm'
assert candidate(994) == 'cmxciv'
assert candidate(1000) == 'm'
# Check some edge cases that are easy to work out by hand.
assert True
| int_to_mini_roman | num = [1, 4, 5, 9, 10, 40, 50, 90,
100, 400, 500, 900, 1000]
sym = ["I", "IV", "V", "IX", "X", "XL",
"L", "XC", "C", "CD", "D", "CM", "M"]
i = 12
res = ''
while number:
div = number // num[i]
number %= num[i]
while div:
res += sym[i]
div -= 1
i -= 1
return res.lower()
| Given a positive integer, obtain its roman numeral equivalent as a string,
and return it in lowercase.
Restrictions: 1 <= num <= 1000
Examples:
>>> int_to_mini_roman(19) == 'xix'
>>> int_to_mini_roman(152) == 'clii'
>>> int_to_mini_roman(426) == 'cdxxvi' | def int_to_mini_roman(number): |
humaneval_157 | def check(candidate):
# Check some simple cases
assert candidate(3, 4, 5) == True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate(1, 2, 3) == False
assert candidate(10, 6, 8) == True
assert candidate(2, 2, 2) == False
assert candidate(7, 24, 25) == True
assert candidate(10, 5, 7) == False
assert candidate(5, 12, 13) == True
assert candidate(15, 8, 17) == True
assert candidate(48, 55, 73) == True
# Check some edge cases that are easy to work out by hand.
assert candidate(1, 1, 1) == False, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate(2, 2, 10) == False
| right_angle_triangle | return a*a == b*b + c*c or b*b == a*a + c*c or c*c == a*a + b*b
| def right_angle_triangle(a, b, c):
'''
Given the lengths of the three sides of a triangle. Return True if the three
sides form a right-angled triangle, False otherwise.
A right-angled triangle is a triangle in which one angle is right angle or
90 degree.
Example:
right_angle_triangle(3, 4, 5) == True
right_angle_triangle(1, 2, 3) == False
''' |
|
humaneval_158 | def check(candidate):
# Check some simple cases
assert (candidate(["name", "of", "string"]) == "string"), "t1"
assert (candidate(["name", "enam", "game"]) == "enam"), 't2'
assert (candidate(["aaaaaaa", "bb", "cc"]) == "aaaaaaa"), 't3'
assert (candidate(["abc", "cba"]) == "abc"), 't4'
assert (candidate(["play", "this", "game", "of","footbott"]) == "footbott"), 't5'
assert (candidate(["we", "are", "gonna", "rock"]) == "gonna"), 't6'
assert (candidate(["we", "are", "a", "mad", "nation"]) == "nation"), 't7'
assert (candidate(["this", "is", "a", "prrk"]) == "this"), 't8'
# Check some edge cases that are easy to work out by hand.
assert (candidate(["b"]) == "b"), 't9'
assert (candidate(["play", "play", "play"]) == "play"), 't10'
| find_max | return sorted(words, key = lambda x: (-len(set(x)), x))[0]
| Write a function that accepts a list of strings.
The list contains different words. Return the word with maximum number
of unique characters. If multiple strings have maximum number of unique
characters, return the one which comes first in lexicographical order.
find_max(["name", "of", "string"]) == "string"
find_max(["name", "enam", "game"]) == "enam"
find_max(["aaaaaaa", "bb" ,"cc"]) == ""aaaaaaa" | def find_max(words): |
humaneval_159 | def check(candidate):
# Check some simple cases
assert True, "This prints if this assert fails 1 (good for debugging!)"
assert candidate(5, 6, 10) == [11, 4], "Error"
assert candidate(4, 8, 9) == [12, 1], "Error"
assert candidate(1, 10, 10) == [11, 0], "Error"
assert candidate(2, 11, 5) == [7, 0], "Error"
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
assert candidate(4, 5, 7) == [9, 2], "Error"
assert candidate(4, 5, 1) == [5, 0], "Error"
| eat | if(need <= remaining):
return [ number + need , remaining-need ]
else:
return [ number + remaining , 0]
| You're a hungry rabbit, and you already have eaten a certain number of carrots,
but now you need to eat more carrots to complete the day's meals.
you should return an array of [ total number of eaten carrots after your meals,
the number of carrots left after your meals ]
if there are not enough remaining carrots, you will eat all remaining carrots, but will still be hungry.
Example:
* eat(5, 6, 10) -> [11, 4]
* eat(4, 8, 9) -> [12, 1]
* eat(1, 10, 10) -> [11, 0]
* eat(2, 11, 5) -> [7, 0]
Variables:
@number : integer
the number of carrots that you have eaten.
@need : integer
the number of carrots that you need to eat.
@remaining : integer
the number of remaining carrots thet exist in stock
Constrain:
* 0 <= number <= 1000
* 0 <= need <= 1000
* 0 <= remaining <= 1000
Have fun :) | def eat(number, need, remaining): |
humaneval_160 | def check(candidate):
# Check some simple cases
assert candidate(['**', '*', '+'], [2, 3, 4, 5]) == 37
assert candidate(['+', '*', '-'], [2, 3, 4, 5]) == 9
assert candidate(['//', '*'], [7, 3, 4]) == 8, "This prints if this assert fails 1 (good for debugging!)"
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| do_algebra | expression = str(operand[0])
for oprt, oprn in zip(operator, operand[1:]):
expression+= oprt + str(oprn)
return eval(expression)
| Given two lists operator, and operand. The first list has basic algebra operations, and
the second list is a list of integers. Use the two given lists to build the algebric
expression and return the evaluation of this expression.
The basic algebra operations:
Addition ( + )
Subtraction ( - )
Multiplication ( * )
Floor division ( // )
Exponentiation ( ** )
Example:
operator['+', '*', '-']
array = [2, 3, 4, 5]
result = 2 + 3 * 4 - 5
=> result = 9
Note:
The length of operator list is equal to the length of operand list minus one.
Operand is a list of of non-negative integers.
Operator list has at least one operator, and operand list has at least two operands. | def do_algebra(operator, operand): |
humaneval_161 | def check(candidate):
# Check some simple cases
assert candidate("AsDf") == "aSdF"
assert candidate("1234") == "4321"
assert candidate("ab") == "AB"
assert candidate("#a@C") == "#A@c"
assert candidate("#AsdfW^45") == "#aSDFw^45"
assert candidate("#6@2") == "2@6#"
# Check some edge cases that are easy to work out by hand.
assert candidate("#$a^D") == "#$A^d"
assert candidate("#ccc") == "#CCC"
# Don't remove this line:
| solve | flg = 0
idx = 0
new_str = list(s)
for i in s:
if i.isalpha():
new_str[idx] = i.swapcase()
flg = 1
idx += 1
s = ""
for i in new_str:
s += i
if flg == 0:
return s[len(s)::-1]
return s
| You are given a string s.
if s[i] is a letter, reverse its case from lower to upper or vise versa,
otherwise keep it as it is.
If the string contains no letters, reverse the string.
The function should return the resulted string.
Examples
solve("1234") = "4321"
solve("ab") = "AB"
solve("#a@C") = "#A@c" | def solve(s): |
humaneval_162 | def check(candidate):
# Check some simple cases
assert candidate('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'
assert candidate('') == None
assert candidate('A B C') == '0ef78513b0cb8cef12743f5aeb35f888'
assert candidate('password') == '5f4dcc3b5aa765d61d8327deb882cf99'
# Check some edge cases that are easy to work out by hand.
assert True
| string_to_md5 | import hashlib
return hashlib.md5(text.encode('ascii')).hexdigest() if text else None
| Given a string 'text', return its md5 hash equivalent string.
If 'text' is an empty string, return None.
>>> string_to_md5('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62' | def string_to_md5(text): |
humaneval_163 | def check(candidate):
# Check some simple cases
assert candidate(2, 10) == [2, 4, 6, 8], "Test 1"
assert candidate(10, 2) == [2, 4, 6, 8], "Test 2"
assert candidate(132, 2) == [2, 4, 6, 8], "Test 3"
assert candidate(17,89) == [], "Test 4"
# Check some edge cases that are easy to work out by hand.
assert True, "This prints if this assert fails 2 (also good for debugging!)"
| generate_integers | lower = max(2, min(a, b))
upper = min(8, max(a, b))
return [i for i in range(lower, upper+1) if i % 2 == 0]
| Given two positive integers a and b, return the even digits between a
and b, in ascending order.
For example:
generate_integers(2, 8) => [2, 4, 6, 8]
generate_integers(8, 2) => [2, 4, 6, 8]
generate_integers(10, 14) => [] | def generate_integers(a, b): |
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