Word Search

Author: Shikha-code36

Backtracking/79-Word-Search.py

@@ -0,0 +1,40 @@
+'''Leetcode- https://leetcode.com/problems/word-search/'''
+'''
+Given an m x n grid of characters board and a string word, return true if word exists in the grid.
+
+The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
+
+Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
+Output: true
+'''
+
+# Solution
+def exist(board, word):
+    ROWS, COLS = len(board), len(board[0])
+    path = set()
+
+    def dfs(r, c, i):
+        if i == len(word):
+            return True
+        if (r < 0 or c < 0 or
+            r >= ROWS or c >= COLS or
+            word[i] != board[r][c] or
+                (r, c) in path):
+            return False
+
+        path.add((r, c))
+        res = (dfs(r+1, c, i+1) or
+               dfs(r-1, c, i+1) or
+               dfs(r, c+1, i+1) or
+               dfs(r, c-1, i+1))
+        path.remove((r, c))
+        return res
+
+    for r in range(ROWS):
+        for c in range(COLS):
+            if dfs(r, c, 0):
+                return True
+    return False
+
+#T: O(n*m*4^n)
+#S: O(n)

README.md

@@ -1 +1,4 @@
-# leetcode-solutions-python
+# leetcode-solutions-python
+
+- [x] [Backtracking](Backtracking)
+    - [x] [Word Search](Backtracking/79-Word-Search.py)