구름용의 개발 블로그

신규 아이디 추천 import re def solution(new_id): allowedChar = '0123456789abcdefghijklmnopqrstuvwxyz-_.' # step 1 new_id = new_id.lower() # step 2 for index, char in enumerate(new_id): if char not in allowedChar: new_id = new_id.replace(char, '') # step 3 new_id = re.sub(r'\.\.+', '.', new_id) # step 4 new_id = new_id.strip('.') # step 5 if new_id == '': new_id = 'a' # step 6 if len(new_id) >= 16: new_..

신고 결과 받기 from collections import defaultdict def solution(id_list, report, k): log = defaultdict(set) answer = [0 for _ in range(len(id_list))] for incident in report: [reporter, reported] = incident.split(' ') log[reported].add(reporter) for id, reporters in log.items(): if len(reporters) < k: continue for reporter in reporters: answer[id_list.index(reporter)] +=1 return answer 정직하게 문제가 요구하는 사항..

def solution(money): # 1번집 털었을 때 dp_A = [0] * len(money) dp_A[0] = money[0] dp_A[1] = max(money[0], money[1]) # 1번 안털고 2번부터 털었을 때 dp_B = [0] * len(money) dp_B[0] = 0 dp_B[1] = money[1] for i in range(2, len(money)-1): dp_A[i] = max(dp_A[i-1], dp_A[i-2] + money[i]) for i in range(2, len(money)): dp_B[i] = max(dp_B[i-1], dp_B[i-2] + money[i]) return max(max(dp_A), max(dp_B))

def solution(m, n, puddles): answer = 0 dp = [[0] * (m+1) for _ in range(n+1)] for puddle in puddles: x,y = puddle dp[y][x] = -1 dp[1][1] = 1 for y in range(1, n+1): for x in range(1, m+1): if x == 1 and y == 1: continue if dp[y][x] == -1: dp[y][x] = 0 else: dp[y][x] = (dp[y-1][x] + dp[y][x-1]) % 1_000_000_007 return dp[n][m]

def solution(triangle): answer = 0 dp = [[] for _ in range(len(triangle))] dp[0].append(triangle[0][0]) for i in range(1, len(triangle)): for index, j in enumerate(triangle[i]): if index == 0: dp[i].append(dp[i-1][0] + j) elif index == len(triangle[i]) - 1: dp[i].append(dp[i-1][-1] + j) else: dp[i].append(max(dp[i-1][index-1] + j, dp[i-1][index] + j)) return max(dp[-1])

from pprint import pprint def solution(N, number): if N == number: return 1 answer = 0 dp = [set() for _ in range(9)] dp[1].add(N) for i in range(2, 9): # N concat dp[i].add(int(str(N) * i)) for j in range(1, i // 2 + 1): for y in dp[j]: for x in dp[i-j]: dp[i].add(x + y) dp[i].add(x - y) dp[i].add(x * y) dp[i].add(-x + y) dp[i].add(-x - y) dp[i].add(-x * y) if y != 0: dp[i].add(x // y) if numbe..

소스 코드 from collections import deque def is_available_change(word, target): cnt = 0 for a1, a2 in zip(word, target): if a1 != a2: cnt += 1 return cnt == 1 def solution(begin, target, words): global answer answer = 100000000000000000000 if target not in words: return 0 visited = [False] * len(words) def dfs(word, depth): global answer if word == target: answer = min(answer, depth) return for index..

소스 코드 def dfs(index, sum, target, numbers): if index == len(numbers): if sum == target: return 1 else: return 0 return dfs(index+1, sum + numbers[index], target, numbers) + dfs(index+1, sum - numbers[index], target, numbers) def solution(numbers, target): answer = 0 answer = dfs(0, 0, target, numbers) return answer print(solution([1,1,1,1,1], 3)) 풀이 위 문제는 백준에서도 본적 있는거 같아서 쉽게 풀었던거 같다. 나올수 있는 정수의 ..

소스 코드 from collections import deque def make_graph(n, wires): # 트리형태를 그래프로 표현 graph = [[] for _ in range(n+1)] for node,neighbor in wires: graph[node].append(neighbor) graph[neighbor].append(node) return graph def solution(n, wires): answer_list = [] graph = make_graph(n, wires) def bfs(graph, start, visited): count = 1 visited[start] = True queue = deque([start]) while queue: node = queue.pople..

소스 코드 def solution(word): word_list = ['A', 'E', 'I', 'O', 'U'] global answer answer = 0 def dfs(_word): global answer answer += 1 if _word == word: return True if len(_word) == 5: return False for string in word_list: if dfs(_word + string): return True for i in word_list: if dfs(i): return answer return answer 풀이 해당 문제는 dfs/bfs를 접해보지 않았기 때문에 기존에는 중복순열을 사용하여 가능한 모든 단어의 조합을 구한 다음 index를 사용하여 몇번째..