In this post, you will know how to solve the Word Search Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Word Search Leetcode SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemGiven 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.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15board 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?Word Search Leetcode Solutions in Pythonclass 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 Solutions in CPPclass 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 Solutions in Javaclass 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 BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Subsets Leetcode Solution Post navigationRemove Duplicates from Sorted Array II Leetcode Solution Subsets Leetcode Solution