HackerRank Making Candies Interview preparation kit problem solution

In this HackerRank Making Candies Interview preparation kit problem you have to complete the minimuPasses function.

Problem solution in Python programming.

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the minimumPasses function below.
def minimumPasses(m, w, p, n):
    days = 0
    candies = 0
    answer = math.ceil(n / (m * w))

    while days < answer:
        if p > candies:
            daysNeeded = math.ceil((p - candies) / (m * w))
            candies += daysNeeded * m * w
            days += daysNeeded

        diff = abs(m - w)
        available = candies // p
        purchased = min(diff, available)

        if m < w:
            m += purchased
        else:
            w += purchased

        rest = available - purchased
        m += rest // 2
        w += rest - rest // 2
        candies -= available * p

        candies += m * w
        days += 1

        remainingCandies = max(n - candies, 0)
        answer = min(answer, days + math.ceil(remainingCandies / (m * w)))
    
    return answer

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

    mwpn = input().split()

    m = int(mwpn[0])

    w = int(mwpn[1])

    p = int(mwpn[2])

    n = int(mwpn[3])

    result = minimumPasses(m, w, p, n)

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

    fptr.close()

Problem solution in Java Programming.

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

public class Solution {

    // Complete the minimumPasses function below.
    static long minimumPasses(long m, long w, long p, long n) {
        long candies = 0;
        long invest = 0;
        long spend = Long.MAX_VALUE;

        while (candies < n) {
                // preventing overflow in m*w
                long passes = (long) (((p - candies) / (double) m) / w);

                if (passes <= 0) {
                        // machines we can buy in total
                        long mw = candies / p + m + w;
                        long half = mw >>> 1;
                        if (m > w) {
                                m = Math.max(m, half);
                                w = mw - m;
                        } else {
                                w = Math.max(w, half);
                                m = mw - w;
                        }
                        candies %= p;
                        passes++;
                }

                // handling overflowing
                // if overflowing is encountered -> candies count are definitely more than long
                // thus it is more than n since n is long
                // so we've reached the goal and we can break the loop
                long mw;
                long pmw;
                try {
                        mw = Math.multiplyExact(m, w);
                        pmw = Math.multiplyExact(passes, mw);
                } catch (ArithmeticException ex) {
                        // we need to add current pass
                        invest += 1;
                        // increment will be 1 because of overflow
                        spend = Math.min(spend, invest + 1);
                        break;
                }

                candies += pmw;
                invest += passes;
                long increment = (long) Math.ceil((n - candies) / (double) mw);
                spend = Math.min(spend, invest + increment);
        }

        return Math.min(spend, invest);
}

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String[] mwpn = scanner.nextLine().split(" ");

        long m = Long.parseLong(mwpn[0]);

        long w = Long.parseLong(mwpn[1]);

        long p = Long.parseLong(mwpn[2]);

        long n = Long.parseLong(mwpn[3]);

        long result = minimumPasses(m, w, p, n);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedWriter.close();

        scanner.close();
    }
}

Problem solution in C++ programming.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

bool check(ll machines, ll workers, ll price, ll target, ll rounds) {
    if (machines >= (target+workers-1)/workers) return true;
    ll cur = machines*workers;
    rounds--;
    if (rounds == 0) return false;
    while (1) {
        ll rem = target - cur;
        ll rnds = (rem + machines*workers - 1) / (machines*workers);
        if (rnds <= rounds) return true;
        if (cur < price) {
          rem = price - cur;
          rnds = (rem + machines*workers - 1) / (machines*workers);
          rounds -= rnds;
          if (rounds < 1) return false;
          cur += rnds * machines * workers;
        }
        cur -= price;
        if (machines > workers) {
          workers++;
        } else {
          machines++;
        }
    }
    return false;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll m, w, p, n;
    cin >> m >> w >> p >> n;
    ll a = 1, b = 1000000000000LL;
    while (a < b) {
        ll mid = (a + b) >> 1;
        if (check(m, w, p, n, mid)) {
          b = mid;
        } else {
          a = mid + 1;
        }
    }
    cout << a << "\n";
    return 0;
}