Question

Data Structure Java Help

Program creates NxN matrix of unique value in correct format and determines if X is located in the matrix in O(N) or better

The input is an N by N matrix of numbers that is already in memory. Each individual
row is increasing from left to right. Each individual column is increasing from
top to bottom. Give an O(N) worst-case algorithm that decides if a number X is in
the matrix.

Answer 1

$\color{blue}\underline{Time \;complexity\;:}$

$\color{blue}\underline{MatrixFind.java:}$

import java.util.*;

public class MatrixFind {
    public static void main(String args[]){
        int N, X;
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter N (size of matrix): ");
        N =  scanner.nextInt();
        Hashtable<Integer,Integer> matrix=new Hashtable<Integer,Integer>();
        System.out.println("Enter matrix (NXN)elements: ");
        for(int i = 0;i<N;i++){
            for(int j = 0;j<N;j++){
                matrix.put(scanner.nextInt(),1);
            }
        }
        System.out.println("Enter key: ");
        X = scanner.nextInt();
        if(matrix.containsKey(X)){
            System.out.println("Number "+ X+" is exists in the matrix.");
        }
        else {
            System.out.println("Number "+ X+" is not exists in the matrix.");
        }
    }
}

$\color{red}\underline{Output:}$

Enter N (size of matrix): Enter matrix (NXN) elements: 1 234 5678 9 10 11 12 13 14 15 16 Enter key: Number 11 is exists in the matrix. Process finished with exit code 0

Enter N (size of matrix): Enter matrix (NXN) elements: 1 234 5678 9 10 11 12 13 14 15 16 Enter key: 17 Number 17 is not exists in the matrix. Process finished with exit code 0