Merge pull request #397 from boostcamp-ai-tech-4/penguin-1211

[펭귄] 2021.12.11

Author: mergify[bot]

coodingpenguin/programmers/level2/짝지어제거하기.py

@@ -0,0 +1,15 @@
+def compress(s):
+ stack = [s[0]]
+ for i in range(1, len(s)):
+ if stack and s[i] == stack[-1]:
+ stack.pop()
+ else:
+ stack.append(s[i])
+ return stack
+
+
+def solution(s):
+ # 압축이 되면
+ if compress(s):
+ return 0 # 0을 반환
+ return 1 # 안 되면 1을 반환

coodingpenguin/programmers/level2/타겟넘버.py

@@ -0,0 +1,16 @@
+def solution(numbers, target):
+ n = len(numbers)
+ count = 0
+ for i in range(1 << n):
+ total = 0
+ for j in range(n):
+ # 비트가 1이면
+ if i & 1 << j:
+ total += numbers[j] # 더해준다
+ # 비트가 0이면
+ else:
+ total -= numbers[j] # 빼준다
+ # 결과가 타겟과 일치하면
+ if total == target:
+ count += 1 # 카운트를 +1 해준다
+ return count

coodingpenguin/programmers/level2/행렬테두리회전하기.py

@@ -0,0 +1,23 @@
+dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
+
+
+def solution(rows, columns, queries):
+ matrix = [[j * columns + i for i in range(1, columns + 1)] for j in range(rows)]
+ result = []
+ for y1, x1, y2, x2 in queries:
+ min_num = 10000 # 최소 숫자
+ cy, cx, cd = y1 - 1, x1 - 1, 0 # 현재 위치(y, x)와 이동 방향
+ new_matrix = [row[:] for row in matrix]
+ # 모든 방향을 돌 때까지 반복
+ while cd < 4:
+ min_num = min(min_num, matrix[cy][cx]) # 최솟값 갱신
+ ny, nx = cy + dirs[cd][0], cx + dirs[cd][1] # 이동할 위치
+ # 범위에서 벗어나면
+ if (ny < y1 - 1 or ny > y2 - 1) or (nx < x1 - 1 or nx > x2 - 1):
+ cd += 1 # 방향을 바꾼다
+ continue
+ new_matrix[ny][nx] = matrix[cy][cx]
+ cy, cx = ny, nx
+ matrix = new_matrix
+ result.append(min_num)
+ return result