add solution: word-search

Author: dusunax

word-search/dusunax.py

@@ -0,0 +1,52 @@
+'''
+# 79. Word Search
+
+use backtracking(DFS) to search for the word in the board.
+
+## Time and Space Complexity
+
+```
+TC: O(n * m * 4^L)
+SC: O(L)
+```
+
+#### TC is O(n * m * 4^L):
+- n is the number of rows in the board.
+- m is the number of columns in the board.
+- L is the length of the word.
+  - 4^L is the number of directions we can go at each step. (explores 4 branches recursively)
+
+#### SC is O(L):
+- modifying the board in-place to mark visited cells. = O(L)
+'''
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        rows = len(board)
+        cols = len(board[0])
+        
+        def backtracking(i, j, word_index): # TC: O(4^L), SC: O(L)
+            if word_index == len(word):
+                return True
+            
+            if i < 0 or i >= rows or j < 0 or j >= cols or board[i][j] != word[word_index]:
+                return False
+            
+            temp = board[i][j]
+            board[i][j] = "."
+            
+            found = (
+                backtracking(i + 1, j, word_index + 1) or
+                backtracking(i - 1, j, word_index + 1) or
+                backtracking(i, j + 1, word_index + 1) or
+                backtracking(i, j - 1, word_index + 1)
+            )
+            board[i][j] = temp
+            
+            return found
+        
+        for row in range(rows): # TC: O(n * m)
+            for col in range(cols):
+                if backtracking(row, col, 0):
+                    return True
+                    
+        return False