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. 

Advantages of Algorithms:

• 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 Problem – Divisible 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();
	}

}