728x90 320x100 💻 하나씩 차곡차곡/프로그래머스 (Python)93 [프로그래머스/Python] [PCCP 모의고사 #1] 1번 - 외톨이 알파벳 def solution(input_string): answer = '' count = {} answer_list = [] # {알파벳 : [알파벳이 있는 인덱스]} for idx, alpha in enumerate(input_string): if alpha not in count: count[alpha] = [idx] else: count[alpha].append(idx) for key, value in count.items(): if len(value) >= 2: for i in range(len(value) - 1): if abs(value[i]-value[i+1]) > 1: answer_list.append(key) break if len(answer_list) == 0: answer = "N" e.. 2024. 1. 18. [프로그래머스/LV.1/Python] 체육복 첫 도전 ... def solution(n, lost, reserve): answer = n - len(lost) for i in range(len(lost)): if n == lost[i]: # 마지막 번호일경우 if lost[i-1] in reserve : answer += 1 reserve.remove(lost[i]-1) elif lost[i] == 1 : # 첫번째 번호일경우 if lost[i+1] in reserve : answer += 1 reserve.remove(lost[i]+1) else : if lost[i]-1 in reserve: answer += 1 reserve.remove(lost[i]-1) elif lost[i]+1 in reserve: answer += 1 reserve.r.. 2024. 1. 17. [프로그래머스/Lv.2/Python] 올바른 괄호 def solution(s): stack = [] for c in s: if c == "(": stack.append(c) else: if stack: # 배열이 비어있지 않으면 stack.pop() else: # 배열이 비여있으면 return False if stack: return False return True 2024. 1. 16. [프로그래머스/Lv.1/Python] 기능개발 import math def solution(progresses, speeds): answer = [] workDay = [] for i in range(len(progresses)): leftPercent = 100 - progresses[i] # 남은 완성률 days = math.ceil(leftPercent / speeds[i]) # 작업일 workDay.append(days) for i in range(len(workDay)): if i != 0 : if workDay[i] < workDay[i-1]: workDay[i] = workDay[i-1] tmp = list(dict.fromkeys(workDay)) for i in tmp : answer.append(workDay.count(i)) re.. 2024. 1. 15. [프로그래머스/Lv.1/Python] 같은 숫자는 싫어 def solution(arr): answer = [] for i in range(len(arr)): if i == 0: # 첫번째 숫자라면 무조건 배열에 넣음. answer.append(arr[i]) else : if answer[-1] != arr[i]: # 연속된 숫자가 아니라면 answer.append(arr[i]) return answer 2024. 1. 12. [프로그래머스/Lv.1/Python] 베스트앨범 def solution(genres, plays): answer = [] total = {} # 장르 : 총재생횟수 genDic = {} # 장르 : (재생횟수, 고유넘버) for i in range(len(genres)): genre = genres[i] play = plays[i] if genres[i] in total.keys(): total[genres[i]] += plays[i] genDic[genres[i]].append((plays[i],i)) else: total[genres[i]] = plays[i] genDic[genre] = [(play,i)] # 가장 재생횟수가 높은 장르 total = sorted(total.items(), key=lambda x: x[1], reverse=True.. 2024. 1. 11. 이전 1 2 3 4 5 ··· 16 다음 728x90 320x100
