word search solution

Author: dolphinflow86

word-search/dolphinflow86.py

@@ -0,0 +1,42 @@
+# 1) Use dfs and backtracking to search for a word in the board.
+# TC: O(N*M*3^W) where N is len(row), M is len(col) and W is len(word)
+# SC: O(N*M) representing the maximum depth of the recursion tack.
+class Solution:
+    def dfs(self, board: List[List[str]], word: str, row: int, col: int, wordIdx: int) -> bool:
+        if row < 0 or col < 0 or row >= len(board) or col >= len(board[0]):
+            return False
+        
+        if wordIdx >= len(word) or board[row][col] != word[wordIdx]:
+            return False
+
+        if wordIdx == len(word) - 1:
+            return True
+        
+        temp = board[row][col]
+        board[row][col] = '#'
+
+        found = (
+            self.dfs(board, word, row + 1, col, wordIdx + 1) or
+            self.dfs(board, word, row - 1, col, wordIdx + 1) or
+            self.dfs(board, word, row, col + 1, wordIdx + 1) or
+            self.dfs(board, word, row, col - 1, wordIdx + 1)
+        )
+
+        board[row][col] = temp
+
+        return found
+
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        n = len(board)
+        m = len(board[0])
+        w = len(word)
+
+        # Early return if the board contains fewer cells than the word length.
+        if n * m < w: return False
+
+        for i in range(n):
+            for j in range(m):
+                if self.dfs(board, word, i, j, 0):
+                    return True
+
+        return False