HackerRank Mark and Toys Interview preparation kit problem solution

In this HackerRank Mark and Toys Interview preparation kit, you have Given a list of toy prices and an amount to spend, determine the maximum number of gifts he can buy.

Problem solution in Python programming.

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the maximumToys function below.
def maximumToys(prices, k):
    items = 0
    prices.sort()
    for p in prices:
        if p <= k:
            items += 1
            k -= p
        else:
            break
    return items

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    nk = input().split()

    n = int(nk[0])

    k = int(nk[1])

    prices = list(map(int, input().rstrip().split()))

    result = maximumToys(prices, k)

    fptr.write(str(result) + '\n')

    fptr.close()

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int numItems = in.nextInt();
        int cash = in.nextInt();

        int[] items = new int[numItems];
        for (int i = 0; i < numItems; i++) {
            items[i] = in.nextInt();
        }

        System.out.println(findNumItemsPurchase(items, cash));
    }

    public static int findNumItemsPurchase(int[] items, int cash) {
        Arrays.sort(items);
        int count = 0;
        for (int i = 0; i < items.length; i++) {
            if (cash - items[i] > 0) {
                cash -= items[i];
                count += 1;
            } else {
                break;
            }
        }

        return count;
    }

}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() 
{
    long long n, k; cin >> n >> k;
    vector<int> prices;
    for(int i = 0; i < n; i++)
    {
        int p; cin >> p;
        prices.push_back(p);
    }
    sort(prices.begin(), prices.end());
    
    int t = 0;
    for(vector<int>::iterator it = prices.begin(); it != prices.end(); it++)
    {
        if(*it <= k)
        {
            t++;
            k -= *it;
        }
        else break;
    }
    cout << t << endl;
    
    return 0;
}

Problem solution in C programming.

#include<stdio.h>
void quicksort(int x[100000],int first,int last){
    int pivot,j,temp,i;
 
     if(first<last){
         pivot=first;
         i=first;
         j=last;
 
         while(i<j){
             while(x[i]<=x[pivot]&&i<last)
                 i++;
             while(x[j]>x[pivot])
                 j--;
             if(i<j){
                 temp=x[i];
                  x[i]=x[j];
                  x[j]=temp;
             }
         }
 
         temp=x[pivot];
         x[pivot]=x[j];
         x[j]=temp;
         quicksort(x,first,j-1);
         quicksort(x,j+1,last);
 
    }}

int main()
{
int n,k,i,avail=0,count=0;
scanf("%d",&n);
scanf("%d",&k);
int cost[n];
for(i=0;i<n;i++)
scanf("%d",&cost[i]); 
quicksort(cost,0,n-1);
while(avail<=k)
{
avail+=cost[count]; 
count++;
}  
printf("%d\n",count-1);    
return 0;    }

Problem solution in JavaScript programming.

var data = '';

var run = function () {
  var parts = data.split('\n'),
    totals = parts[0].split(' '),
    numToys = +totals[0],
    money = +totals[1],
    toyPrices = parts[1].split(' ').sort(function (a, b) { return a - b; }),
    total = 0,
    i = 0;

    while (total < money) {
      total += +toyPrices[i++];
    }

    console.log(--i);
}


process.stdin.resume();
process.stdin.setEncoding('ascii');
process.stdin.on('data', function (input) {
  data += input;
});

process.stdin.on('end', run);