HackerRank Mark and Toys problem solution

In this HackerRank Mark and Toys Interview preparation kit problem solution, Mark and Jane are very happy after having their first child. Their son loves toys, so Mark wants to buy some. There are a number of different toys lying in front of him, tagged with their prices.

Mark has only a certain amount to spend, and he wants to maximize the number of toys he buys with this money. Given a list of toy prices and an amount to spend, determine the maximum number of gifts he can buy.

Function Description

Complete the function maximumToys in the editor below.

maximumToys has the following parameter(s):

• int prices[n]: the toy prices
• int k: Mark’s budget

Returns

• int: the maximum number of toys

HackerRank Mark and Toys problem solution in Python.

#!/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()

Mark and Toys problem solution in Java.

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("%dn",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);