[Leetcode] 0079. Word Search

Explore diverse LeetCode solutions in Python, C++, JavaScript, SQL, and TypeScript. Ideal for interview prep, learning, and code practice in multiple programming languages. Github Repo Link

Given an xn 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.

Example 1:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E ","E"]], word = "ABCCED"
Output: true

Example 2:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E ","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E ","E"]], word = "ABCB"
Output: false

Constraints:

m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

Table of contents

Python

				
					# time complexity: O(c*3^l) # space complexity: O(l) from typing import List class Solution: def exist(self, board: List[List[str]], word: str) -> bool: def backtrack(suffix: str, r: int, c: int): if len(suffix) == 0: return True if not (0 <= r < ROW and 0 <= c < COL) or suffix[0] ! = board[r][c]: return False result = False originalChar = board[r][c] board[r][c] = "#" for dr,dc in ([1,0],[0,1 ],[-1,0],[0,-1]): result = backtrack(suffix[1:], r+dr, c+dc) if result: break board[r][c] = originalChar return result ROW = len(board) COL = len(board[0]) for row in range(ROW): for col in range(COL): if backtrack(word, row, col): return True return False board = [[" A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E" ]] word = "ABCCED" print(Solution().exist(board, word))

Python

related articles