HackerRank Luck Balance Problem Solution
In this post, we will solve HackerRank Luck Balance Problem 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, L[i] and T[i]:
L[i] is the amount of luck associated with a contest. If Lena wins the contest, her luck balance will decrease by L[i]: if she loses it, her luck balance will increase by L[i].
T[i] denotes the contest’s importance rating. It’s equal to 1 if the contest is important, and it’s equal to 0 if it’s unimportant.
If Lena loses no more than k 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
k = 2
L = [5, 1, 4]
T= [1,1,0]
Contest L[i] T[i]
1 5 1
2 1 1
3 4 0If Lena loses all of the contests, her will be 5+1+4 = 10. Since she is allowed to lose 2 important contests, and there are only 2 important contests, she can lose all three contests to maximize her luck at 10.
If k = 1, she has to win at least 1 of the 2 important contests. She would choose to win the lowest value important contest worth 1. Her final luck will be 5+ 4-1 = 8.
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 contests[i] contains two integers that represent the luck balance and importance of the ¿th contest
Returns
- int: the maximum luck balance achievable
Input Format
The first line contains two space-separated integers n and k, the number of preliminary contests and the maximum number of important contests Lena can lose.
Each of the next n lines contains two space-separated integers, L[i] and T[i], the contest’s luck balance and its importance rating.
ample 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
29Explanation
There are n = 6 contests. Of these contests, 4 are important and she cannot lose more thank = 3 of them. Lena maximizes her luck if she wins the 3rd important contest (where Li]= 1) and loses all of the other five contests for a total luck balance of 5+2+8+10+5-1=29.
Luck Balance C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int cmp(const void *a,const void *b)
{
return *(int *)b-*(int *)a;
}
int main() {
int n,k,i,j=0,m=0,c;
long int sum=0;
long int b,sum1=0;
scanf("%d%d",&n,&k);
int a[n+1];
for(i=0;i<n;i++)
{
scanf("%ld%d",&b,&c);
if(c==1)
{
a[j++]=b;
sum1+=b;
}
else if(c==0)
sum1+=b;
}
if(j>=1)
qsort(a,j,sizeof(int),cmp);
for(i=k;i<j;i++)
sum1-=(2*a[i]);
printf("%ld\n",sum1);
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}Luck Balance C++ Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n, k, l, t;
cin >> n >> k;
vector<int> important;
int ans = 0;
for (int i = 0; i < n; ++i) {
cin >> l >> t;
ans += (1 - t) * l;
if (t) {
important.push_back(l);
}
}
sort(important.begin(), important.end());
for (int i = 0; i < (int) important.size(); ++i) {
int coef = -1;
if ((int)important.size() - i <= k) coef *= coef;
ans += important[i] * coef;
}
cout << ans << endl;
return 0;
}Luck Balance C++ Solution
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {
static void Main(String[] args) {
string[] paramters = Console.ReadLine().Split(' ');
int N = Convert.ToInt32(paramters[0]);
int K = Convert.ToInt32(paramters[1]);
List<Contest> importantContests = new List<Contest>();
List<Contest> unimportantContests = new List<Contest>();
for (int i = 0; i < N; i++)
{
string input = Console.ReadLine();
var contest = new Contest(input);
if (contest.Important)
{
importantContests.Add(contest);
}
else
{
unimportantContests.Add(contest);
}
}
if (importantContests.Count < K)
{
Console.WriteLine(importantContests.Sum(x => x.Luck) + unimportantContests.Sum(x => x.Luck));
}
else
{
var orderedImportantContests = importantContests.OrderByDescending(x => x.Luck);
int luckGained = orderedImportantContests.Take(K).Sum(x => x.Luck) + unimportantContests.Sum(x => x.Luck);
int luckLost = orderedImportantContests.Skip(K).Take(importantContests.Count - K).Sum(x => x.Luck);
Console.WriteLine(luckGained - luckLost);
}
}
}
public class Contest
{
public int Luck { get; set; }
public bool Important { get; set; }
public Contest(string input)
{
string[] values = input.Split(' ');
this.Luck = Convert.ToInt32(values[0]);
this.Important = Convert.ToInt32(values[1]) == 1;
}
}Luck Balance Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Solution {
public static void main(String args[]) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int k=sc.nextInt();
int values[]=new int[n];
int imp[]=new int[n];
int l=-1,m=-1;
for(int i=0;i<n*2;i++){
if(i%2!=0)
imp[++l]=sc.nextInt();
else
values[++m]=sc.nextInt();
}
for(int i=0;i<n-1;i++)
for(int j=i+1;j<n;j++){
if(values[j]>values[i]){
int temp=values[i];
int temp2=imp[i];
values[i]=values[j];
values[j]=temp;
imp[i]=imp[j];
imp[j]=temp2;
}
}
int Result=0;
for(int i=0;i<n;i++){
if(imp[i]==1&&k>0){
Result+=values[i];
k--;
}
else if(imp[i]==1&&k==0)
Result-=values[i];
else
Result+=values[i];
}
System.out.println(Result);
}
}Luck Balance JavaScript Solution
function processData(input) {
//Enter your code here
var inputAsLines = input.split('\n');
var N = parseInt(inputAsLines[0].split(' ')[0]);
var K = parseInt(inputAsLines[0].split(' ')[1]);
var runningTotal = 0;
var impArray = new Array();
for(var i=0;i<N;i++) {
var line = inputAsLines[i+1].split(' ');
var Li = parseInt(line[0]);
var Ti = parseInt(line[1]);
runningTotal+=Li;
if (Ti==1)
impArray.push(Li);
}
impArray.sort(function(a, b) {
return a - b;
});
var outputArray = impArray.slice(0,impArray.length - K);
var output = outputArray.reduce(function(pv, cv) { return pv + cv; }, 0);
if (K>=impArray.length) {
console.log(runningTotal);
}
else
{
console.log(runningTotal - output*2);
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});Luck Balance Python Solution
n, k = input().strip().split()
n, k = int(n), int(k)
totalLuck = 0
importantLucks = []
importantLuckCount = 0
for i in range(n):
li, ti = input().strip().split()
if ti == "0":
totalLuck += int(li)
else:
importantLucks.append(int(li))
importantLuckCount += 1
importantLucks = sorted(importantLucks, reverse = True)
while importantLuckCount > k:
totalLuck -= importantLucks[importantLuckCount-1]
importantLucks = importantLucks[: (importantLuckCount-1)]
importantLuckCount -= 1
for l in importantLucks:
totalLuck += l
print(totalLuck)
