HackerRank Making Candies problem solution YASH PAL, 31 July 202410 September 2024 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; } coding problems interview prepration kit