Key Pair [EASY]

Company: Microsoft, Flipkart, Carwale, Amazon

Question: 

Given an array A[] of n numbers and another number x, determine whether or not there exist two elements in A whose sum is exactly x.

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 X,N is the size of array.
The second line of each test case contains N integers representing array elements C[i].

Output:

Print "Yes" if there exist two elements in A whose sum is exactly x, else "No" without quotes.

Constraints:

1 ≤ T ≤ 200
1 ≤ N ≤ 200
1 ≤ C[i] ≤ 1000

Example:

Input:
2
6 16
1 4 45 6 10 8
5 10
1 2 4 3 6

Output:
Yes
Yes

CODE:

import java.util.*;

import java.lang.*;

import java.io.*;

class Solution {

public static void main (String[] args) {

Scanner sc = new Scanner(System.in);

int m = sc.nextInt();

for(int mi=0;mi<m;mi++){

    int size= sc.nextInt();

    int tot= sc.nextInt();

    int count=0;

    int [] ar= new int [size];

    for(int i=0;i<size;i++){

        ar[i]=sc.nextInt();

    }

    for(int j=0;j<size;j++){

        for(int k =0;k<size;k++){

            if(ar[j]+ar[k]==tot){count++;}

        }

    }

    if(count>=1){System.out.println("Yes");}

    else{System.out.println("No");}

}

}

}

EXECUTION TIME:0.21s

CODE B: Using Binary Hash-Map via a boolean array 

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

class GFG {
public static void main (String[] args) {
Scanner sc = new Scanner (System.in);
    int m = sc.nextInt();
    for(int mi=0;mi<m;mi++){
     int n1=sc.nextInt();
     int n2=sc.nextInt();
     int []a1= new int[n1];
     for(int a1_i=0;a1_i<n1;a1_i++){
         a1[a1_i]=sc.nextInt();
     }
     check(a1,n1,n2);
    }
}
public static void check(int [] a, int n,int sum){
   boolean [] binmap= new boolean[100000];
   int count=0;
   for(int i=0;i<n;i++){
       int temp=sum-a[i];
       if(temp>=0 && binmap[temp]){
           count++;
       }
       binmap[a[i]]=true;
   }
   if(count>=1){System.out.println("Yes");}
   else{System.out.println("No");}
}

}

EXECUTION TIME:0.19s