230502 - deque 활용

[robert-min]

괄호 회전하기 링크

• 풀이 : stack, deque 활용

  def solution(s):
  answer = 0
  # deque
  from collections import deque
  n = len(s)
  q = deque(list(s))
  while n > 0:
      stack = list()
      for i in q:
          if i in ["[", "{", "("]:
              stack.append(i)
          # "]", "}", ")" 인경우
          else:
              if not stack:
                  stack.append(i)
                  break
              else:  
                  if i == "]" and stack[-1] == "[":
                      stack.pop()
                  elif i == "}" and stack[-1] == "{":
                      stack.pop()
                  elif i == ")" and stack[-1] == "(":
                      stack.pop()
                  else:
                      stack.append(i)
                      break
      if not stack:
          answer += 1
  
      q.append(q.popleft())
      n -= 1    
  return answer

[robert-min]

캐시 링크

• 풀이 : cacheSize == 0 일때, 예외처리

  def solution(cacheSize, cities):
  answer = 0
  from collections import deque
  cache = deque([])
  for city in cities:
      city = city.lower()
      if city in cache:
          cache.remove(city)
          cache.append(city)
          answer += 1
      else:                
          if len(cache) < cacheSize:
              cache.append(city)
          elif cacheSize != 0:
              cache.popleft()
              cache.append(city)
          answer += 5
  
  return answer

[robert-min]

n^2 배열 링크

• 정답풀이

  def solution(n, left, right):
  return [max(i//n, i%n)+1 for i in range(left, right+1)]

• 시간 초과 풀이

  def solution(n, left, right):
  answer = []
  # n * n 행렬
  graph = [[1] * n for _ in range(n)]
  for i in range(2, n+1):
      for j in range(i):
          graph[i-1][j] = i
          graph[j][i-1] = i
  
  temp = []
  for idx, g in enumerate(graph):
      temp.extend(g)
  return temp[left:right+1]