079 Word Search

Author: idf

079 Word Search py3.py

@@ -0,0 +1,63 @@
+#!/usr/bin/python3
+"""
+Given a 2D board and a word, find if the word exists in the grid.
+
+The word can be constructed from letters of sequentially adjacent cell, where
+"adjacent" cells are those horizontally or vertically neighboring. The same
+letter cell may not be used more than once.
+
+Example:
+
+board =
+[
+  ['A','B','C','E'],
+  ['S','F','C','S'],
+  ['A','D','E','E']
+]
+
+Given word = "ABCCED", return true.
+Given word = "SEE", return true.
+Given word = "ABCB", return false.
+"""
+from typing import List
+from collections import defaultdict
+
+
+dirs = [(0, -1), (0, 1), (1, 0), (-1, 0)]
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        m, n = len(board), len(board[0])
+        visited = defaultdict(lambda: defaultdict(bool))
+        for i in range(m):
+            for j in range(n):
+                if board[i][j] == word[0]:
+                    if self.dfs(board, visited, i, j, word, 1):
+                        return True
+
+        return False
+
+    def dfs(self, board, visited, i, j, word, idx):
+        visited[i][j] = True
+        if idx >= len(word):
+            return True
+
+        m, n = len(board), len(board[0])
+        for di, dj in dirs:
+            I = i + di
+            J = j + dj
+            if 0 <= I < m and 0 <= J < n and not visited[I][J] and board[I][J] == word[idx]:
+                if self.dfs(board, visited, I, J, word, idx + 1):
+                    return True
+
+        visited[i][j] = False  # restore
+        return False
+
+
+if __name__ == "__main__":
+    assert Solution().exist([
+        ["A","B","C","E"],
+        ["S","F","E","S"],
+        ["A","D","E","E"]
+    ], "ABCESEEEFS") == True