Word Search Leetcode Solution

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.

Word Search Leetcode Solution
Word Search Leetcode Solution

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 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?

Now, lets 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.

NEXT: Remove Duplicates from Sorted Array II

Sharing Is Caring

Leave a Comment