Chocolate Feast – HackerRank Solution

In this post, we will solve Chocolate Feast HackerRank Solution. This problem (Chocolate Feast) is a part of HackerRank Problem Solving series.

Task

Little Bobby loves chocolate. He frequently goes to his favorite 5 & 10 store, Penny Auntie, to buy them. They are having a promotion at Penny Auntie. If Bobby saves enough wrappers, he can turn them in for a free chocolate.

Example

n = 15
c = 3
m = 2

He has 15 to spend, bars cost 3, and he can turn in 2 wrappers to receive another bar. Initially, he buys 5 bars and has 5 wrappers after eating them. He turns in 4 of them, leaving him with 1, for 2 more bars. After eating those two, he has 3 wrappers, turns in 2 leaving him with 1 wrapper and his new bar. Once he eats that one, he has 2 wrappers and turns them in for another bar. After eating that one, he only has  wrapper, and his feast ends. Overall, he has eaten 5 + 2 + 1 + 1 = 9 bars.

Function Description

Complete the chocolateFeast function in the editor below.

chocolateFeast has the following parameter(s):

• int n: Bobby’s initial amount of money
• int c: the cost of a chocolate bar
• int m: the number of wrappers he can turn in for a free bar

Returns

• int: the number of chocolates Bobby can eat after taking full advantage of the promotion

Note: Little Bobby will always turn in his wrappers if he has enough to get a free chocolate.

Input Format

The first line contains an integer, t, the number of test cases to analyze.
Each of the next t lines contains three space-separated integers: n, c, and m. They represent money to spend, cost of a chocolate, and the number of wrappers he can turn in for a free chocolate.

Constraints

• 1 <= t <= 1000
• 2 <= n <= 10⁵
• 1 <= c <= n
• 2 <= m <= n

Sample Input

STDIN   Function
-----   --------
3       t = 3 (test cases)
10 2 5  n = 10, c = 2, m = 5 (first test case)
12 4 4  n = 12, c = 4, m = 4 (second test case)
6 2 2   n = 6,  c = 2, m = 2 (third test case)

Sample Output

6
3
5

Explanation

Bobby makes the following 3 trips to the store:

1. He spends 10 on 5 chocolates at 2 apiece. He then eats them and exchanges all 5 wrappers to get 1 more. He eats 6 chocolates.
2. He spends his 12 on 3 chocolates at 4 apiece. He has 3 wrappers, but needs 4 to trade for his next chocolate. He eats 3 chocolates.
3. He spends 6 on 3 chocolates at 2 apiece. He then exchanges 2 of the 3 wrappers for 1 additional piece. Next, he uses his third leftover chocolate wrapper from his initial purchase with the wrapper from his trade-in to do a second trade-in for 1 more piece. At this point he has 1 wrapper left, which is not enough to perform another trade-in. He eats 5 chocolates.

Solution – Chocolate Feast – HackerRank Solution

C++

#include <iostream>
using namespace std;

int main() {
    int t, n, c, m;
    cin >> t;

    while (t--) {
        cin >> n >> c >> m;
        int answer = n / c;
        int wrappers = answer;

        // Spend wrappers for additional chocolates.
        while (wrappers >= m) {
            wrappers -= m;

            // Eat chocolate, produce new wrapper.
            answer++;
            wrappers++;
        }

        cout << answer << endl;
    }

    return 0;
}

Python

import sys

def wrappers(wr, m):
    res = 0
    if wr//m > 0:
        res += wr//m + wrappers(wr//m + wr%m, m)
    return res
    

def chocolateFeast(n, c, m):
    return n//c + wrappers(n//c, m)


if __name__ == "__main__":
    t = int(input().strip())
    for a0 in range(t):
        n, c, m = input().strip().split(' ')
        n, c, m = [int(n), int(c), int(m)]
        result = chocolateFeast(n, c, m)
        print(result)

Java

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

public class Solution {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        scanner.nextLine();
        while (scanner.hasNext()) {
            int money = scanner.nextInt();
            int price = scanner.nextInt();
            int bonus = scanner.nextInt();
            
            int count = money / price;
            int wrappers = count;
            
            while (wrappers >= bonus) {
                int freebies = wrappers / bonus;
                count += freebies;
                wrappers = freebies + wrappers % bonus;
            }
            
            System.out.println(count);
        }
    }
}

Note: This problem (Chocolate Feast) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.