Count pair with given sum [EASY-Hashmap]-Very Important

Given an array of integers, and an integer  ‘K’ , find the count of pairs of elements in the array whose sum is equal to 'K'.

Input:
First line of the input contains an integer T, denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. First line of each test case contains 2 space separated integers N and K denoting the size of array and the sum respectively. Second line of each test case contains N space separated integers denoting the elements of the array.

Output:
Print the count of pairs of elements in the array whose sum is equal to the K.

Constraints:
1<=T<=50
1<=N<=50
1<=K<=50
1<=A[i]<=100

Example:
Input
2
4 6
1  5  7 1
4 2
1 1 1 1
Output
2
6
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 n=sc.nextInt();
     int sum=sc.nextInt();
     int [] a= new int[n];
     for(int j=0;j<n;j++){
         a[j]=sc.nextInt();
     }
     System.out.println(Countpairs(a,sum,n));
 }
}
public static int Countpairs(int[] arr, int sum, int n){
   HashMap<Integer, Integer> hm = new HashMap<>();

        // Store counts of all elements in map hm
        for (int i=0; i<n; i++){
             
            // initializing value to 0, if key not found
            if(!hm.containsKey(arr[i]))
                hm.put(arr[i],0);
                 
            hm.put(arr[i], hm.get(arr[i])+1);
        }
        int twice_count = 0;

        // iterate through each element and increment the
        // count (Notice that every pair is counted twice)
        for (int k=0; k<n; k++)
        {
            if(hm.get(sum-arr[k]) != null)
                twice_count += hm.get(sum-arr[k]);

            // if (arr[i], arr[i]) pair satisfies the condition,
            // then we need to ensure that the count is
            // decreased by one such that the (arr[i], arr[i])
            // pair is not considered
            if (sum-arr[k] == arr[k])
                twice_count--;
        }

        // return the half of twice_count
        return (twice_count/2);
}

}

EXECUTION TIME: 0.11s