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https://school.programmers.co.kr/learn/courses/30/lessons/155652 파이썬 def solution(s, skip, index): text = [each for each in 'abcdefghijklmnopqrstuvwxyz' * 3 if each not in skip] return ''.join(text[text.index(each) + index] for each in s) 테스트 1 〉통과 (0.02ms, 10.1MB) 테스트 2 〉통과 (0.02ms, 10.2MB) 테스트 3 〉통과 (0.02ms, 10.3MB) 테스트 4 〉통과 (0.01ms, 10.4MB) 테스트 5 〉통과 (0.02ms, 10.2MB) 테스트 6 〉통과 (0.02ms, 10.2M..

https://school.programmers.co.kr/learn/courses/30/lessons/150370 파이썬 짧게 줄인다면.. def solution(today, terms, privacies): def sum_date(year, month, day, t_month): q, month = divmod(month + t_month, 12) if month = sum_date(*(int(each) for each in p_date.split('.')), terms[t_num])] 읽기 편하게 코드를 작성한다면... def solution(today, terms, privacies): def str2date(text): return [int(each) for each in text.split('..

https://school.programmers.co.kr/learn/courses/30/lessons/147355 파이썬 solution = lambda t, p: sum(1 for i in range(len(t) - len(p) + 1) if int(t[i: i + len(p)])

https://school.programmers.co.kr/learn/courses/30/lessons/142086 파이썬 def solution(s): marked = {} answer = [] for index, letter in enumerate(s): answer.append(index - marked[letter] if letter in marked else -1) marked[letter] = index return answer 코틀린 class Solution { fun solution(s: String): IntArray { val answer = mutableListOf() val counter = mutableMapOf() for ((index, each) in s.withIndex()..

https://school.programmers.co.kr/learn/courses/30/lessons/140108 파이썬 def solution(s): first = s[0] new_index = slices = same = no_same = 0 for index, each in enumerate(s): new_index += 1 if each == first: same += 1 else: no_same += 1 if same == no_same: slices += 1 new_index = same = no_same = 0 if index + 1 < len(s): first = s[index + 1] return slices + (1 if new_index else 0) def solution(s): ..

https://school.programmers.co.kr/learn/courses/30/lessons/138477 파이썬 힙큐를 이용하면 조금 빠르려나... def solution(k, score): import heapq answer, honer = [], [] for each in score: heapq.heappush(honer, each) if len(honer) > k: heapq.heappop(honer) answer.append(honer[0]) # honer[0] = 가장 작은 원소 return answer 테스트 1 〉통과 (0.01ms, 10MB) 테스트 2 〉통과 (0.01ms, 10.1MB) 테스트 3 〉통과 (0.01ms, 10.1MB) 테스트 4 〉통과 (0.01ms, 9.97..

https://school.programmers.co.kr/learn/courses/30/lessons/136798 파이썬 약수는 제곱근(루트)를 이용해서 계산을 줄일 수 있다. def solution(number, limit, power): answer = 0 for i in range(1, number + 1): sqrt_i = i ** .5 count = sum(1 for j in range(1, int(sqrt_i) + 1) if i % j == 0) * 2 - (0 if sqrt_i % 1 else 1) answer += power if limit < count else count return answer 코틀린 class Solution { fun solution(number: Int, lim..

https://school.programmers.co.kr/learn/courses/30/lessons/135808 파이썬 for 문을 이용. def solution(k, m, score): score.sort(reverse=True) return sum(score[index] * m for index in range(m - 1, len(score), m)) slice를 이용. def solution(k, m, score): return sum(sorted(score, reverse=True)[m - 1:: m]) * m 코틀린 class Solution { fun solution(k: Int, m: Int, score: IntArray): Int { val sortedScore = score.sorte..