💁🏻 나의 풀이 from collections import deque def solution(bridge_length, weight, truck_weights): queue = deque(truck_weights) bridge = deque([0] * bridge_length) bridge.popleft() head = queue.popleft() bridge.append(head) answer = 1 count = 1 total = head while total > 0 or queue: head = bridge.popleft() if head > 0: total -= head count -= 1 if queue and queue[0] + total

💁🏻 나의 풀이 import heapq def solution(N, road, K): answer = 0 graph = {node: {} for node in range(1, N+1)} distances = {node: int(2000 * 1e4) for node in range(1, N+1)} distances[1] = 0 queue = [] heapq.heappush(queue, [distances[1], 1]) for path in road: a, b, c = path if b in graph[a] and c >= graph[a][b]: continue graph[a][b] = c graph[b][a] = c while queue: distance_now, destination_now = heapq..

💁🏻 나의 풀이 def solution(clothes): answer = 1 items = {} for item, category in clothes: if category not in items: items[category] = [] items[category].append(item) for k in items.keys(): answer *= len(items[k])+1 return answer-1 종류를 key로, 리스트를 item으로 갖는 dict에 모든 옷을 넣음 종류별로 최대 하나씩만 입을 수 있기 때문에 각 종류별 옷의 개수 + 1 를 모두 곱한 값 - 1 각 종류별 옷의 개수 + 1 = 각 옷을 입는 경우 + 하나도 안입는 경우 모두 곱한 값 - 1 = 모든 경우의 수 - 하나도 안입는 경우..

💁🏻 나의 풀이 from collections import deque def is_correct(s): stack = deque() for c in s: stack.append(c) if len(stack) > 1: if stack[-2] + stack[-1] in ['[]', '{}', '()']: stack.pop() stack.pop() return len(stack) == 0 def solution(s): answer = 0 s = deque(list(s)) for i in range(len(s)): s.rotate(1) if is_correct(s): answer += 1 return answer from collections import deque def is_correct(s): stack ..

💁🏻 나의 풀이 from itertools import combinations def is_unique(rel, c): unique = set() for row in rel: arr = [] for i in c: arr.append(row[i]) unique.add(tuple(arr)) return len(rel) == len(unique) def solution(relation): answer = 0 column_cnt = len(relation[0]) comb = [] for i in range(1, column_cnt + 1): comb.extend(combinations(range(column_cnt), i)) unique_comb = [] for c in comb: if is_unique(rel..