HackerRank Marc’s Cakewalk problem solution

HackerRank Marc’s Cakewalk problem solution – Marc loves cupcakes, but he also likes to stay fit. Each cupcake has a calorie count, and Marc can walk a distance to expend those calories. If Marc has eaten j cupcakes so far, after eating a cupcake with c calories he must walk at least 2^j x c miles to maintain his weight.

Given the individual calorie counts for each of the cupcakes, determine the minimum number of miles Marc must walk to maintain his weight. Note that he can eat the cupcakes in any order.

Function Description

Complete the marcsCakewalk function in the editor below.

marcsCakewalk has the following parameter(s):

• int calorie[n]: the calorie counts for each cupcake

HackerRank Marc’s Cakewalk problem solution in Python.

#!/bin/python3

import sys


n = int(input().strip())
cal = list(map(int, input().strip().split(' ')))
# your code goes here
cal.sort()
cal.reverse()

miles=0
for i in range(n):
    cup=cal[i]*(2**i)
    miles+=cup
    
print(miles)

Marc’s Cakewalk problem solution in Java.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        Integer[] calories = new Integer[n];
        for(int calories_i=0; calories_i < n; calories_i++){
            calories[calories_i] = in.nextInt();
        }
        List<Integer> calList = Arrays.asList(calories);
        Collections.sort(calList, Collections.reverseOrder());

        long cals = 0;
        for (int i=0;i<calList.size();i++) {
            cals += Math.pow(2,i)*calList.get(i);
        }
        System.out.println(cals);
    }
}

Problem solution in C++.

#include <bits/stdc++.h>

using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> calories(n);
    for(int calories_i = 0; calories_i < n; calories_i++){
       cin >> calories[calories_i];
    }
    // your code goes here
    sort(calories.begin(),calories.end());
    reverse(calories.begin(),calories.end());
    long long temp=1,ans=0;
    for(int i=0;i<n;i++)
    {
        ans+=calories[i]*temp;
        temp*=2;
    }
    printf("%lldn",ans);
    return 0;
}

Problem solution in C.

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int n; 
    scanf("%d",&n);
    int a[n];
    int i,j,t;
    long int c=0;
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
        for(i=0;i<n;i++){
            for(j=n-1;j>0;j--){
                if(a[j]>a[j-1]){
                    t=a[j];
                    a[j]=a[j-1];
                    a[j-1]=t;
                }
            }
        }
        for(int i=0;i<n;i++){
            c=c+(a[i]*(pow(2,i)));
        }
    printf("%ld",c);
    return 0;
}