Given an **m x n** **grid of characters** board and a **string word**, return** true if the 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.

**Examples:**

Example 1:Input:[ ["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"] ] word = "ABCCED"Output:trueExplanation:We can easily find the given word in the matrix.Example 2:Input:[ ["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"] ] word = "ABCB"Output:falseExplanation:There is no such word in the given matrix.

**Solution:**

**Disclaimer:** *Don’t jump directly to the solution, try it out yourself first.*

**Prerequisite: **You should be aware of backtracking. If not then follow Backtracking (Basics to Advanced)

**Approach:** We are going to solve this by using backtracking, in this approach first we will linearly search the entire matrix to find the first letters matching our given string. If we found those letters then we can start backtracking in all four directions to find the rest of the letters of the given string.

**Step 1:** Find the first character of the given string.

**Step 2:** Start Backtracking in all four directions until we find all the letters of sequentially adjacent cells.

**Step 3:** At the end, If we found our result then return true else return false.

**Edge cases:** Now think about what will be our stopping condition, we can stop or return false if we **reach the end of the boundaries **of the matrix or the letter at which we are making recursive calls **is not the required letter**.

We will also return if **we found all the letters** of the given word i.e. we found the number of letters equal to the length of the given word.

*NOTE: Do not forget that we cannot reuse a cell again.*

That is, we have to somehow keep track of our position so that we cannot find the same letter again and again.

In this approach, we are going to mark visited cells with some random character that will prevent us from revisiting them again and again.

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
bool searchNext(vector<vector<char>> &board, string word, int row, int col,
int index, int m, int n) {
// if index reaches at the end that means we have found the word
if (index == word.length())
return true;
// Checking the boundaries if the character at which we are placed is not
//the required character
if (row < 0 || col < 0 || row == m || col == n || board[row][col] !=
word[index] or board[row][col] == '!')
return false;
// this is to prevent reusing of the same character
char c = board[row][col];
board[row][col] = '!';
// top direction
bool top = searchNext(board, word, row - 1, col, index + 1, m, n);
// right direction
bool right = searchNext(board, word, row, col + 1, index + 1, m, n);
// bottom direction
bool bottom = searchNext(board, word, row + 1, col, index + 1, m, n);
// left direction
bool left = searchNext(board, word, row, col - 1, index + 1, m, n);
board[row][col] = c; // undo change
return top || right || bottom || left;
}
bool exist(vector<vector<char>> board, string word) {
int m = board.size();
int n = board[0].size();
int index = 0;
// First search the first character
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == word[index]) {
if (searchNext(board, word, i, j, index, m, n))
return true;
}
}
}
return false;
}
int main() {
vector<vector<char>> board {{'A','B','C','E'},
{'S','F','C','S'},
{'A','D','E','E'}};
string word = "ABCCED";
bool res = exist(board, word);
if(res==1)
cout<<"True"<<endl;
else
cout<<"False"<<endl;
}
```

**Output:**True

**Time Complexity: O(m*n*4^k)**, *where “K” is the length of the word. And we are searching for the letter m*n times in the worst case. Here 4 in 4^k is because at each level of our decision tree we are making 4 recursive calls which equal 4^k in the worst case.*

**Space Complexity: O(K)**, *Where k is the length of the given words.*

## Java Code

```
import java.io.*;
import java.lang.*;
class Solution {
public static void main(String[] args) {
char[][] board = {{'A','B','C','E'},
{'S','F','C','S'},
{'A','D','E','E'}};
String word = "ABCCED";
Solution sol = new Solution();
boolean res = sol.exist(board, word);
System.out.println(res);
}
public boolean exist(char[][] board, String word) {
int m = board.length;
int n = board[0].length;
int index = 0;
// First search the first character
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == word.charAt(index)) {
if (searchNext(board, word, i, j, index, m, n))
return true;
}
}
}
return false;
}
private boolean searchNext(char[][] board, String word, int row, int col,
int index, int m, int n) {
// if index reaches at the end that means we have found the word
if (index == word.length())
return true;
// Checking the boundaries if the character at which we are placed is not
//the required character
if (row < 0 || col < 0 || row == m || col == n || board[row][col] !=
word.charAt(index) || board[row][col] == '!')
return false;
// this is to prevent reusing of the same character
char c = board[row][col];
board[row][col] = '!';
// top direction
boolean top = searchNext(board, word, row - 1, col, index + 1, m, n);
// right direction
boolean right = searchNext(board, word, row, col + 1, index + 1, m, n);
// bottom direction
boolean bottom = searchNext(board, word, row + 1, col, index + 1, m, n);
// left direction
boolean left = searchNext(board, word, row, col - 1, index + 1, m, n);
board[row][col] = c; // undo change
return top || right || bottom || left;
}
}
```

**Output:**True

**Time Complexity: O(m*n*4^k)**, *where “K” is the length of the word. And we are searching for the letter m*n times in the worst case. Here 4 in 4^k is because at each level of our decision tree we are making 4 recursive calls which equal 4^k in the worst case.*

**Space Complexity: O(K)**, *Where k is the length of the given words.*

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