Mark and Toys

Problem Statement :

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


int: the maximum number of toys
Input Format

The first line contains two integers,  and , the number of priced toys and the amount Mark has to spend.
The next line contains  space-separated integers

Solution :


                            Solution in C :

In  C  :

void quicksort(int x[100000],int first,int last){
    int pivot,j,temp,i;

int main()
int n,k,i,avail=0,count=0;
int cost[n];
return 0;    }

                        Solution in C++ :

In  C++ :

#include <string>
#include <vector>
#include <map>
#include <list>
#include <iterator>
#include <set>
#include <queue>
#include <iostream>
#include <sstream>
#include <stack>
#include <deque>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <utility> 
using namespace std;
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define RFOR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define REP(i, N) FOR(i, 0, N)
#define RREP(i, N) RFOR(i, N, 0)
#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define Pi 3.14159265358979

typedef long long Int;
typedef unsigned long long UInt;
typedef vector <int> VI;
typedef pair <int, int> PII;

int a[1<<17], n, k;

clock_t startTime;

int main()
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	startTime = clock();
	Int sum = 0;
	int res = 0;
	while (sum <= k)
		sum += a[res];
		if (sum > k)
	printf("%d\n", res);
//	fprintf(stderr, "Used time: %f", (clock() - startTime) / (CLOCKS_PER_SEC*1.0));
	return 0;

                        Solution in Java :

In  Java :

import java.util.*;

public class Solution{		
		private BufferedReader in;	
		private StringTokenizer st;
		private PrintWriter out;
		void solve() throws IOException{
			int n = nextInt();
			int k = nextInt();
			int []x = new int[n];
			for (int i = 0; i < x.length; i++) {
				x[i] = nextInt();
			long sum = 0;
			int ans = 0;
			for (int i = 0; i < x.length; i++) {
				sum += x[i];
				if(sum <= k){

		Solution() throws IOException {
			in = new BufferedReader(new InputStreamReader(;	
			out = new PrintWriter(System.out);

		private void eat(String str) {
			st = new StringTokenizer(str);

		String next() throws IOException {
			while (!st.hasMoreTokens()) {
				String line = in.readLine();				
				if (line == null) {					
					return null;
			return st.nextToken();

		int nextInt() throws IOException {
			return Integer.parseInt(next());

		long nextLong() throws IOException {
			return Long.parseLong(next());

		double nextDouble() throws IOException {
			return Double.parseDouble(next());

		public static void main(String[] args) throws IOException {
			new Solution();

		int gcd(int a,int b){
			if(b>a) return gcd(b,a);
			if(b==0) return a;
			return gcd(b,a%b);


                        Solution in Python : 
In  Python3 :

x = input ()
parts = x.split ()
part1 = int (parts [0])
part2 = int (parts [1])
y = input ()
arr = []
y = y.split ()
for i in range (0, part1):
	arr.append (int (y [i]))
arr.sort ()
i = 0
j = 0
while j <part1 and part2> arr [j]:
	part2- = arr [j]
	j + = 1
	i + = 1
print (i)

View More Similar Problems


This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →

Jesse and Cookies

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

View Solution →

Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

View Solution →

Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

View Solution →

Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

View Solution →

Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

View Solution →