geeksforgeeks@ Equal to product (Binary Search)

Posted 2020-08-03 流白

http://www.practice.geeksforgeeks.org/problem-page.php?pid=667

Equal to product

Given an array of integers check whether there are two numbers present with given product.

Input:

The first line of input contains an integer T denoting the number of test cases.
The first line of each test case is N and a product p.
The second line of each test case contain N number of a[].

Output:
Print Yes if two numbers product is equal to p else No.

Constraints:

1 ≤ T ≤ 51
1 ≤ N ≤ 100
0 ≤ a[] ≤ 1000
1 ≤ pro ≤ 2000000

Example:

Input:

2

5 2

1 2 3 4 5

8 46

5 7 9 22 15 344 92 8

Output:

Yes

No

 

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG {
    
    public static boolean check(int[] arr, int q) {
        
        Arrays.sort(arr);
        int n = arr.length;
        
        for(int i=0; i<n; ++i) {
            if(arr[i] == 0) {
                if(q == 0) return true;
                else continue;
            }
            int remain = q % arr[i];
            if(remain != 0) continue;
            
            int div = q / arr[i];
            int pos = Arrays.binarySearch(arr, div);
            if(pos >= 0) {
                return true;
            }
        }
        return false;
    }
    
    public static void main (String[] args) {
        Scanner in = new Scanner(System.in);
        int times = in.nextInt();
        
        while(times > 0) {
            int n = in.nextInt();
            int q = in.nextInt();
            int[] arr = new int[n];
            
            for(int i=0; i<n; ++i) {
                arr[i] = in.nextInt();
            }
            
            boolean rs = check(arr, q);
            System.out.println(rs? "Yes": "No");
            --times;
        }
    }
}

View Code