Approximately II Codechef Solution

Hello Programmers In this post, you will know how to solve the Approximately II Codechef Solution. The Problem Code: APPROX2

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Codechef by yourself. If you find any difficulty after trying several times, then you can look for solutions.

Problem

You are given an array of N integers a₁, a₂, …, a_N and an integer K. Find the number of such unordered pairs {i, j} that

i ≠ j
|a_i + a_j – K| is minimal possible

Output the minimal possible value of |a_i + a_j – K| (where i ≠ j) and the number of such pairs for the given array and the integer K.

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case consists of two space separated integers – N and K respectively.
The second line contains N single space separated integers – a₁, a₂, …, a_N respectively.

Output

For each test case, output a single line containing two single space separated integers – the minimal possible value of |a_i + a_j – K| and the number of unordered pairs {i, j} for which this minimal difference is reached.

Constraints

1 ≤ T ≤ 50
1 ≤ a_i, K ≤ 10⁹
N = 2 – 31 point.
2 ≤ N ≤ 1000 – 69 points.

Example

Sample Input 1

1
4 9
4 4 2 6

Sample Output 1

1 4

report this adApproximately II CodeChef Solution in JAVA

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for (int tc = 0; tc < T; tc++) {
			int N = sc.nextInt();
			int K = sc.nextInt();
			int[] a = new int[N];
			for (int i = 0; i < a.length; i++) {
				a[i] = sc.nextInt();
			}
			System.out.println(solve(a, K));
		}
		sc.close();
	}
	static String solve(int[] a, int K) {
		int minDiff = Integer.MAX_VALUE;
		int pairCount = 0;
		for (int i = 0; i < a.length; i++) {
			for (int j = i + 1; j < a.length; j++) {
				int diff = Math.abs(a[i] + a[j] - K);
				if (diff < minDiff) {
					minDiff = diff;
					pairCount = 1;
				} else if (diff == minDiff) {
					pairCount++;
				}
			}
		}
		return String.format("%d %d", minDiff, pairCount);
	}
}