**Unidirectional Word Search - Amazon Top Interview Questions**

### Problem Statement :

Given a two-dimensional matrix of characters board and a string target, return whether the target can be found in the matrix by going left-to-right, or up-to-down unidirectionally. Constraints n, m ≤ 250 where n is the number of rows and columns in board k ≤ 250 where k is the length of word Example 1 Input board = [ ["H", "E", "L", "L", "O"], ["A", "B", "C", "D", "E"] ] word = "HELLO" Output True Example 2 Input board = [ ["x", "z", "d", "x"], ["p", "g", "u", "x"], ["k", "j", "z", "d"] ] word = "xgz" Output False Explanation You can't make "xgz" going left-to-right or up-to-down.

### Solution :

` ````
Solution in C++ :
bool solve(vector<vector<string>>& board, string word) {
int n = board.size();
int m = board[0].size();
vector<vector<char>> adj(n);
char ch = word[0];
queue<pair<int, int>> q;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
adj[i].push_back(board[i][j][0]);
if (board[i][j][0] == ch) q.push({i, j});
}
}
while (!q.empty()) {
int i = q.front().first;
int j = q.front().second;
q.pop();
int len = 0;
for (int k = j; k < m; k++) {
if (adj[i][k] == word[len])
len++;
else
break;
}
if (len == word.length()) return true;
len = 0;
for (int k = i; k < n; k++) {
if (adj[k][j] == word[len])
len++;
else
break;
}
if (len == word.length()) return true;
}
return false;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(String[][] board, String word) {
for (int i = 0; i < board.length; i++) {
StringBuilder sbr = new StringBuilder();
for (int j = 0; j < board[0].length; j++) sbr.append(board[i][j]);
if (sbr.toString().contains(word))
return true;
}
for (int i = 0; i < board[0].length; i++) {
StringBuilder sbr = new StringBuilder();
for (int j = 0; j < board.length; j++) sbr.append(board[j][i]);
if (sbr.toString().contains(word))
return true;
}
return false;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, board, word):
r = len(board)
c = len(board[0])
for i in range(c):
k = 0
for j in range(r):
if board[j][i] == word[k]:
k += 1
if k == len(word):
return True
for b in board:
p = 0
for i, c in enumerate(b):
if c == word[p]:
p += 1
if p == len(word):
return True
return False
```

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