[PGS] 프렌즈4블록 / Level 2 / 1시간 / 실패

Author: Mingguriguri

minjeong/Simulation/2025-07-05-[PGS]-프렌즈4블록.py

@@ -1,63 +1,45 @@
-def check_remove(m, n, grid):
-    """
-    2×2 같은 블록을 찾아서, 지울 칸을 True로 표시한 2D 배열을 반환한다.
-    """
-    mark = [[False] * n for _ in range(m)]
-    for i in range(m-1):
-        for j in range(n-1):
-            current = grid[i][j]
-            if current and grid[i][j+1] == current \
-                and grid[i+1][j] == current \
-                and grid[i+1][j+1] == current:
-                mark[i][j] = mark[i][j+1] = mark[i+1][j] = mark[i+1][j+1] = True
-    return mark
-
-def remove_blocks(m, n, grid, mark):
-    """
-    mark 배열이 True인 칸으르 찾아서 빈칸으로 바꾸고,
-    삭제된 블록의 총 개수를 반환한다.
-    """
-    removed = 0
+def check(m, n, board):
+    filter = [[0 for _ in range(n)] for _ in range(m)]
+    count = 0
+
+    for i in range(m - 1):
+        for j in range(n - 1):
+            a = board[i][j]
+            b = board[i][j + 1]
+            c = board[i + 1][j]
+            d = board[i + 1][j + 1]
+            if a == b == c == d and a != '0':
+                filter[i][j], filter[i][j + 1], filter[i + 1][j], filter[i + 1][j + 1] = 1, 1, 1, 1
+
     for i in range(m):
         for j in range(n):
-            if mark[i][j]:
-                grid[i][j] = ""
-                removed += 1
-    return removed
-
-def down_blocks(m, n, grid):
-    """
-    중력을 적용하여, 각 열마다 빈칸 위의 블록을 아래로 당긴다.
-    """
-    for j in range(n):
-        stack = []
-        for i in range(m):
-            if grid[i][j] != "":
-                stack.append(grid[i][j])
-
-        for i in range(m-1, -1, -1):
-            if stack:
-                grid[i][j] = stack.pop()
-
-            else:
-                grid[i][j] = ""
+            if filter[i][j] == 1:
+                count += 1
+                board[i][j] = '0'
+
+    if count == 0:
+        return 0
+
+    for i in range(m - 2, -1, -1):
+        for j in range(n):
+            k = i
+            while 0 <= k + 1 < m and board[k + 1][j] == '0':
+                k += 1
+            if k != i:
+                board[k][j] = board[i][j]
+                board[i][j] = '0'
+
+    return count
+
 
 def solution(m, n, board):
-    # 1. 입력 문자열을 2D 리스트로 변환
-    grid = [list(row) for row in board]
-    total_removed = 0
+    answer = 0
+    board = list(map(list, board))
 
     while True:
-        # 2. 지울 칸 표시
-        mark = check_remove(m, n, grid)
-        # 3. 표시된 칸 삭제 및 개수 세기
-        removed = remove_blocks(m, n, grid, mark)
-        if removed == 0:
-            # 더이상 지울 블록이 없다면 반복 종료
+        temp = check(m, n, board)
+        if temp == 0:
             break
-        # 4. 전체 삭제 개수에 해당 칸의 삭제 개수 누적해서 더하기
-        total_removed += removed
-        # 5. 중력 적용
-        down_blocks(m, n, grid)
+        answer += temp
 
-    return total_removed
+    return answer