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Counter6

0118. Ordered Count of Characters Count the number of occurrences of each character and return it as a (list of tuples) in order of appearance. For empty output return (an empty list). Consult the solution set-up for the exact data structure implementation depending on your language. Example: ordered_count("abracadabra") == [('a', 5), ('b', 2), ('r', 2), ('c', 1), ('d', 1)] Solution: def ordered_count(inp): from collections impo.. 2023. 1. 18.
1023. No Loops 2 - You only need one No Loops Allowed You will be given an array a and a value x. All you need to do is check whether the provided array contains the value, without using a loop. Array can contain numbers or strings. x can be either. Return true if the array contains the value, false if not. With strings you will need to account for case. Looking for more, loop-restrained fun? Check out the other kata in the series:.. 2022. 10. 23.
Find the stray number You are given an odd-length array of integers, in which all of them are the same, except for one single number. Complete the method which accepts such an array, and returns that single different number. The input array will always be valid! (odd-length >= 3) Examples [1, 1, 2] ==> 2 [17, 17, 3, 17, 17, 17, 17] ==> 3 Solution: def stray(arr): import collections return [k for k, v in collections.C.. 2022. 8. 19.
Find the unique number There is an array with some numbers. All numbers are equal except for one. Try to find it! find_uniq([ 1, 1, 1, 2, 1, 1 ]) == 2 find_uniq([ 0, 0, 0.55, 0, 0 ]) == 0.55 It’s guaranteed that array contains at least 3 numbers. The tests contain some very huge arrays, so think about performance. Solution: def find_uniq(arr): n = arr[0] == arr[-1] and arr[0] or arr[1] return [i for i in arr if n != i.. 2022. 8. 8.
Vowel Count Description: Return the number (count) of vowels in the given string. We will consider a, e, i, o, u as vowels for this Kata (but not y). The input string will only consist of lower case letters and/or spaces. Solution: 1. Check whether the characters in the 'sentence' are included in 'aeiou'. 2. Returns the number of included characters. def get_count(sentence): return sum([i in 'aeiou' for i i.. 2022. 4. 22.
문자열 내 p와 y의 개수 문제 설명 대문자와 소문자가 섞여있는 문자열 s가 주어집니다. s에 'p'의 개수와 'y'의 개수를 비교해 같으면 True, 다르면 False를 return 하는 solution를 완성하세요. 'p', 'y' 모두 하나도 없는 경우는 항상 True를 리턴합니다. 단, 개수를 비교할 때 대문자와 소문자는 구별하지 않습니다. 예를 들어 s가 "pPoooyY"면 true를 return하고 "Pyy"라면 false를 return합니다. 제한 사항 문자열 s의 길이 : 50 이하의 자연수 문자열 s는 알파벳으로만 이루어져 있습니다. 해결 방법 1. s를 대문자로 만들어준다. 2. P와 Y의 개수를 각각 구한다. 3. 개수가 서로 일치하는지 확인한다. def solution(s): s = s.upper() p_cn.. 2022. 2. 2.