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79\. Word Search

Medium

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:**

![](https://assets.leetcode.com/uploads/2020/11/04/word2.jpg)

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

**Output:** true 

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg)

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

**Output:** true 

**Example 3:**

![](https://assets.leetcode.com/uploads/2020/10/15/word3.jpg)

**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`?

To solve the "Word Search" problem in Java with the Solution class, follow these steps:

1. Define a method `exist` in the `Solution` class that takes a 2D character array `board` and a string `word` as input and returns `true` if the `word` exists in the `board`.
2. Implement a backtracking algorithm to search for the `word` in the `board`.
3. Iterate through each cell in the `board`:
   - For each cell, call a recursive helper function `search` to check if the `word` can be found starting from that cell.
   - If `search` returns `true`, return `true` immediately.
4. Define the `search` method to perform the recursive backtracking:
   - Check if the current cell is out of bounds or if the current character in the `board` does not match the corresponding character in the `word`.
   - If any of the conditions are met, return `false`.
   - Mark the current cell as visited by changing its value to a special character (e.g., `#`) to avoid revisiting it.
   - Recursively call `search` on neighboring cells (up, down, left, right) with the next character in the `word`.
   - After exploring all possible paths from the current cell, backtrack by restoring the original value of the current cell.
5. If the `search` method reaches the end of the `word`, return `true`.
6. If no match is found after exploring all cells, return `false`.

Here's the implementation of the `exist` method in Java:

```java
class Solution {
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (search(board, word, i, j, 0))
                    return true;
            }
        }
        return false;
    }
    
    private boolean search(char[][] board, String word, int i, int j, int index) {
        if (index == word.length())
            return true;
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != word.charAt(index))
            return false;
        char temp = board[i][j];
        board[i][j] = '#'; // Mark as visited
        boolean found = search(board, word, i + 1, j, index + 1) ||
                        search(board, word, i - 1, j, index + 1) ||
                        search(board, word, i, j + 1, index + 1) ||
                        search(board, word, i, j - 1, index + 1);
        board[i][j] = temp; // Restore original value
        return found;
    }
}
```

This implementation uses backtracking to search for the word in the board, with a time complexity of O(M * N * 4^L), where M and N are the dimensions of the board and L is the length of the word.