Luck Balance 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 ProblemLuck Balance – Hacker Rank Solution

Luck Balance – Hacker Rank Solution


Lena is preparing for an important coding competition that is preceded by a number of sequential preliminary contests. Initially, her luck balance is 0. She believes in “saving luck”, and wants to check her theory. Each contest is described by two integers,  and :

  •  is the amount of luck associated with a contest. If Lena wins the contest, her luck balance will decrease by ; if she loses it, her luck balance will increase by .
  •  denotes the contest’s importance rating. It’s equal to  if the contest is important, and it’s equal to  if it’s unimportant.

If Lena loses no more than  important contests, what is the maximum amount of luck she can have after competing in all the preliminary contests? This value may be negative.


Contest		L[i]	T[i]
1		5	1
2		1	1
3		4	0

If Lena loses all of the contests, her will be . Since she is allowed to lose  important contests, and there are only  important contests, she can lose all three contests to maximize her luck at .

If , she has to win at least  of the  important contests. She would choose to win the lowest value important contest worth . Her final luck will be .

Function Description

Complete the luckBalance function in the editor below.

luckBalance has the following parameter(s):

  • int k: the number of important contests Lena can lose
  • int contests[n][2]: a 2D array of integers where each  contains two integers that represent the luck balance and importance of the  contest


  • int: the maximum luck balance achievable

Input Format

The first line contains two space-separated integers  and , the number of preliminary contests and the maximum number of important contests Lena can lose.
Each of the next  lines contains two space-separated integers,  and , the contest’s luck balance and its importance rating.


Sample Input

STDIN       Function
-----       --------
6 3         n = 6, k = 3
5 1         contests = [[5, 1], [2, 1], [1, 1], [8, 1], [10, 0], [5, 0]]
2 1
1 1
8 1
10 0
5 0

Sample Output



There are  contests. Of these contests,  are important and she cannot lose more than  of them. Lena maximizes her luck if she wins the  important contest (where ) and loses all of the other five contests for a total luck balance of .

Luck Balance– Hacker Rank Solution
import java.util.Scanner;

 * @author Techno-RJ
public class LuckBalance {
	public static void sort(int a[], int lo, int hi) {
		if (hi > lo) {
			int q = partition(a, lo, hi);
			sort(a, lo, q - 1);
			sort(a, q + 1, hi);

	public static int partition(int a[], int lo, int hi) {
		int i = lo;
		int j = hi + 1;
		while (hi > lo) {

			while (a[++i] < a[lo])
				if (i == hi)
			while (a[--j] > a[lo])
				if (j == lo)

			if (i >= j)
			swap(a, i, j);

		swap(a, lo, j);
		return j;

	public static void swap(int a[], int i, int j) {
		int temp = a[i];
		a[i] = a[j];
		a[j] = temp;

	public static void main(String[] args) {
		Scanner sc = new Scanner(;
		int N = sc.nextInt();
		int K = sc.nextInt();
		int win = 0;
		int a[] = new int[N];
		int sum = 0;
		for (int i = 0; i < N; i++) {
			int temp = sc.nextInt();
			if (sc.nextInt() == 1) {
				a[i] = temp;
			} else {
				a[i] = Integer.MAX_VALUE;
			sum += temp;
		sort(a, 0, a.length - 1);
		int s2 = 0;
		for (int i = 0; i < win - K; i++) {
			s2 += a[i];

		System.out.println(sum - 2 * s2);

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