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 :



title-img 




                        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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