# Divisible Sum Pairs in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct]

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#### Introduction To Algorithm

The word Algorithm means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results.

• It is easy to understand.
• Algorithm is a step-wise representation of a solution to a given problem.
• In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

Link for the ProblemDivisible Sum Pairs– Hacker Rank Solution

`Divisible Sum Pairs – Hacker Rank Solution`

### Problem:

Function Description

Complete the divisibleSumPairs function in the editor below.

divisibleSumPairs has the following parameter(s):

• int n: the length of array
• int ar[n]: an array of integers
• int k: the integer divisor

Returns
– int: the number of pairs

Input Format

The first line contains 2 space-separated integers, n and k.
The second line contains n space-separated integers, each a value of arr[i].

Constraints

Sample Input

```STDIN           Function
-----           --------
6 3             n = 6, k = 3
1 3 2 6 1 2     ar = [1, 3, 2, 6, 1, 2]
```

Sample Output

``` 5
```

Explanation

`Divisible Sum Pairs – Hacker Rank Solution`
```import java.util.Scanner;

/**
* @author Techno-RJ
*
*/
public class DivisibleSumPairs {

static int divisibleSumPairs(int n, int k, int[] ar) {
int[] bucket = new int[k];
int pairCounter = 0;

for (int element : ar) {

int remainder = element % k;
int complement = (k-remainder)%k;
pairCounter += ar[complement];
bucket[remainder]++;

}
return pairCounter;
}

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();

int k = sc.nextInt();
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
//		int count = 0;
//		for (int i = 0; i < M; i++) {
//			for (int j = i + 1; j < M; j++) {
//				if ((a[i] + a[j]) % N == 0) {
//					count++;
//				}
//			}
//		}
int count = divisibleSumPairs(n, k, a);
System.out.println(count);
sc.close();
}

}```  