[최원준] Day20

leetcode3/최원준/1034. Coloring A Border.py

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+#
+
+'''
+1. 아이디어 :
+border인 애들을 구해야한다. border의 조건은 grid의 끝에 위치하거나, 인접한 4개가 모두 같은 숫자가 아닌 경우.
+dfs를 통해 border인지 구분하여 candids에 넣는다.
+candids을 하나씩 순회하며 grid를 업데이트한다.
+
+2. 시간복잡도 :
+    O(n * m * (4+1))
+
+3. 자료구조/알고리즘 :
+dfs
+
+'''
+class Solution:
+    def colorBorder(self, grid: List[List[int]], row: int, col: int, color: int) -> List[List[int]]:
+        n = len(grid)
+        m = len(grid[0])
+
+        base_num = grid[row][col]
+        candids = []
+        dx = [0,0,1,-1]
+        dy = [1,-1,0,0]
+        visited = [[0] * m for _ in range(n)]
+
+        def dfs(x, y):
+            if visited[x][y]:
+                return
+
+            visited[x][y] = 1
+            if grid[x][y] != base_num:
+                return
+
+            is_border = False
+            adjacent = 0
+            for i in range(4):
+                nx = x + dx[i]
+                ny = y + dy[i]
+                if not (0 <= nx < n and 0 <= ny < m):
+                    is_border = True
+                    break
+                if grid[nx][ny] == base_num:
+                    adjacent += 1
+
+            if is_border or adjacent < 4:
+                candids.append([x,y])
+
+            for i in range(4):
+                nx = x + dx[i]
+                ny = y + dy[i]
+                if not (0<=nx<n and 0<=ny<m):
+                    continue
+                dfs(nx, ny)
+
+        dfs(row, col)
+
+        for x, y in candids:
+            grid[x][y] = color
+
+        return grid
+
+
+
+
+
+        # [1,1] 0 0 3
+        # [1,2]
+
+        # [3,3]
+        # [3,2]
+
+        # [1,2,2] 0 1 3
+        # [2,3,2]
+
+        # [1,3,3]
+        # [2,3,3]
+
+        # [1,1,1] 1 1 2
+        # [1,1,1]
+        # [1,1,1]
+
+        # [2,2,2]
+        # [2,1,2]
+        # [2,2,2]

leetcode3/최원준/3838. Weighted Word Mapping.py

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+#
+
+'''
+1. 아이디어 :
+-
+
+2. 시간복잡도 :
+    O(n * m) n = words.length , m =words[i].length
+
+3. 자료구조/알고리즘 :
+-
+
+'''
+class Solution:
+    def mapWordWeights(self, words: List[str], weights: List[int]) -> str:
+        return "".join(chr(ord('z') - sum(weights[ord(c) - ord('a')] for c in word) %26 ) for word in words)