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In this post, you will find the solution for **Luck Balance** **in Java-HackerRank Problem**. We are providing the **correct and tested solutions** of coding problems present on **HackerRank**. If you are not able to solve any problem, then you can take help from our Blog/website.

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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** – Luck Balance – Hacker Rank Solution

Luck Balance – Hacker Rank Solution

**Problem:**

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.

**Example**

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

**Returns**

*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.

**Constraints**

**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**

29

**Explanation**

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) break; while (a[--j] > a[lo]) if (j == lo) break; if (i >= j) break; 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(System.in); 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) { win++; 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); sc.close(); } }