HackerRank Sherlock and The Beast problem solution

In this HackerRank Sherlock and The Beast problem solution Sherlock Holmes suspects his archenemy Professor Moriarty is once again plotting something diabolical. Sherlock’s companion, Dr. Watson, suggests Moriarty may be responsible for MI6’s recent issues with their supercomputer, The Beast.

Shortly after resolving to investigate, Sherlock receives a note from Moriarty boasting about infecting The Beast with a virus. He also gives him a clue: an integer. Sherlock determines the key to removing the virus is to find the largest Decent Number having that number of digits.

A Decent Number has the following properties:

1. Its digits can only be 3’s and/or 5’s.
2. The number of 3’s it contains is divisible by 5.
3. The number of 5’s it contains is divisible by 3.
4. It is the largest such number for its length.

Moriarty’s virus shows a clock counting down to The Beast’s destruction, and time is running out fast. Your task is to help Sherlock find the key before The Beast is destroyed!

Problem solution in Python.

def decent_number(N):
    """
    Prints a decent number of N digits. -1 if no such number exists.
    """
    
    for fives in range(N, -1, -1):
        threes = N - fives
        if fives % 3 == 0 and threes % 5 == 0:
            return "5" * fives + "3" * threes
    return "-1"
    
def main():
    
    import sys
    
    count, *Ns = sys.stdin.readlines()
    
    for N in Ns:
        print(decent_number(int(N)))
    
if __name__ == "__main__":
    main()
    

Problem solution in Java.

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

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int count = scan.nextInt();
        int piv;
        for (int x = 0; x < count; x++) {
            piv = scan.nextInt();
            decent(piv, pivot(piv));
        }
    }
    
    private static void decent(int n, int piv) {
        if (piv < 0) {
            System.out.println("-1");
        }
        else {
            int repeat = piv/3;
            while(repeat-- > 0)
                System.out.print("555");
            repeat = (n - piv)/5;
            while(repeat-- > 0)
                System.out.print("33333");
            System.out.println();
        }
    }
    
    private static int pivot(int n) {
        while(n > 0) {
            if(n % 3 == 0)
                break;
            else
                n -= 5;
        }
        return n;
    }
}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int T;
    scanf("%d",&T);
    while (T--)
    {
        int n;
        scanf("%d",&n);
        int ans5=-1;
        for (int i=0;i*3<=n;i++)
        {
            int tot=n-3*i;
            if (tot%5==0)
                ans5=3*i;
        }
        if (ans5==-1) puts("-1");
        else
        {
            for (int i=0;i<ans5;i++)
                printf("5");
            for (int i=0;i<n-ans5;i++)
                printf("3");
            puts("");
        }
    }
    return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int T; int n;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        int mfive = -1;
        for(int five = 0; five * 3 <= n; five++){
            if((n - five * 3)%5 == 0)
                mfive = five;
        }
        if(mfive == -1)
            printf("%dn", -1);
        else{
            for(int i = 0; i < mfive * 3; i++)
                putchar('5');
            for(int i = mfive * 3; i < n; i++)
                putchar('3');
            puts("");
        }
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}