In this post, we are going to solve the Word Search Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
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.
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
andword
consists of only lowercase and uppercase English letters.
Follow up: Could you use search pruning to make your solution faster with a larger board
?
Now, let’s see the leetcode solution of Word Search Leetcode Solution.
Word Search Leetcode Solution in Python
class Solution: def exist(self, board: List[List[str]], word: str) -> bool: m = len(board) n = len(board[0]) def dfs(i: int, j: int, s: int) -> bool: if i < 0 or i == m or j < 0 or j == n: return False if board[i][j] != word[s] or board[i][j] == '*': return False if s == len(word) - 1: return True cache = board[i][j] board[i][j] = '*' isExist = \ dfs(i + 1, j, s + 1) or \ dfs(i - 1, j, s + 1) or \ dfs(i, j + 1, s + 1) or \ dfs(i, j - 1, s + 1) board[i][j] = cache return isExist return any(dfs(i, j, 0) for i in range(m) for j in range(n))
Word Search Leetcode Solution in CPP
class Solution { public: bool exist(vector<vector<char>>& board, string word) { for (int i = 0; i < board.size(); ++i) for (int j = 0; j < board[0].size(); ++j) if (dfs(board, word, i, j, 0)) return true; return false; } private: bool dfs(vector<vector<char>>& board, const string& word, int i, int j, int s) { if (i < 0 || i == board.size() || j < 0 || j == board[0].size()) return false; if (board[i][j] != word[s] || board[i][j] == '*') return false; if (s == word.length() - 1) return true; const char cache = board[i][j]; board[i][j] = '*'; const bool isExist = dfs(board, word, i + 1, j, s + 1) || dfs(board, word, i - 1, j, s + 1) || dfs(board, word, i, j + 1, s + 1) || dfs(board, word, i, j - 1, s + 1); board[i][j] = cache; return isExist; } };
Word Search Leetcode Solution in Java
class Solution { public boolean exist(char[][] board, String word) { for (int i = 0; i < board.length; ++i) for (int j = 0; j < board[0].length; ++j) if (dfs(board, word, i, j, 0)) return true; return false; } private boolean dfs(char[][] board, String word, int i, int j, int s) { if (i < 0 || i == board.length || j < 0 || j == board[0].length) return false; if (board[i][j] != word.charAt(s) || board[i][j] == '*') return false; if (s == word.length() - 1) return true; final char cache = board[i][j]; board[i][j] = '*'; final boolean isExist = dfs(board, word, i + 1, j, s + 1) || dfs(board, word, i - 1, j, s + 1) || dfs(board, word, i, j + 1, s + 1) || dfs(board, word, i, j - 1, s + 1); board[i][j] = cache; return isExist; } }
Note: This problem Word Search is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
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