HackerRank Beautiful Days at the Movies Solution

Yashwant Parihar,

In this post, we will solve HackerRank Beautiful Days at the Movies Problem. Lily likes to play games with integers. She has created a new game where she determines the difference between a number and its reverse. For instance, given the number 12, its reverse is 21. Their difference is 9. The number 120 reversed is 21, and their difference is 99.

She decides to apply her game to decision-making. She will look at a numbered range of days and will only go to a movie on a beautiful day. Given a range of numbered days, [i ……. ĵ] and a number k, determine the number of beautiful days in the field. Beautiful numbers are defined as numbers where i-reverse(i) is evenly divisible by k. If a day’s value is a beautiful number, it is a beautiful day. Return the number of beautiful days in the range.

Function Description

Complete the beautifulDays function in the editor below.

beautiful days has the following parameter(s):

  • int i: the starting day number
  • int j: the ending day number
  • int k: the divisor

Returns

  • int: the number of beautiful days in the range

Input Format

A single line of three space-separated integers describing the respective values of i, j, and k.

Sample Input

20 23 6

Sample Output

2
HackerRank Beautiful Days at the Movies Solution
HackerRank Beautiful Days at the Movies Solution

Beautiful Days at the Movies C Solution

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

int reverse(int n) {
    int rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n /= 10;
    }
    return rev;
}

int beautifulDays(int i, int j, int k) {
    int count = 0;
    for (int day = i; day <= j; day++) {
        int rev = reverse(day);
        if (abs(day - rev) % k == 0) {
            count++;
        }
    }
    return count;
}

int main() {
    int i, j, k;
    scanf("%d %d %d", &i, &j, &k);
    int result = beautifulDays(i, j, k);
    printf("%d\n", result);
    return 0;
}

Beautiful Days at the Movies C++ Solution

#include <iostream>
#include <cmath>

using namespace std;

int reverse(int n) {
    int rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n /= 10;
    }
    return rev;
}

int main() {
    int i, j, k, beautiful_days = 0;
    cin >> i >> j >> k;
    for (int day = i; day <= j; day++) {
        int rev = reverse(day);
        if (abs(day - rev) % k == 0) {
            beautiful_days++;
        }
    }
    cout << beautiful_days << endl;
    return 0;
}

Beautiful Days at the Movies C Sharp Solution

using System;

class Solution {
    static int Reverse(int n) {
        int rev = 0;
        while (n > 0) {
            rev = rev * 10 + n % 10;
            n /= 10;
        }
        return rev;
    }

    static void Main(String[] args) {
        string[] tokens_i = Console.ReadLine().Split(' ');
        int i = int.Parse(tokens_i[0]);
        int j = int.Parse(tokens_i[1]);
        int k = int.Parse(tokens_i[2]);
        int beautiful_days = 0;
        for (int day = i; day <= j; day++) {
            int rev = Reverse(day);
            if (Math.Abs(day - rev) % k == 0) {
                beautiful_days++;
            }
        }
        Console.WriteLine(beautiful_days);
    }
}

Beautiful Days at the Movies Java Solution

import java.util.Scanner;

public class Solution {
    static int reverse(int n) {
        int rev = 0;
        while (n > 0) {
            rev = rev * 10 + n % 10;
            n /= 10;
        }
        return rev;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int i = in.nextInt();
        int j = in.nextInt();
        int k = in.nextInt();
        int beautiful_days = 0;
        for (int day = i; day <= j; day++) {
            int rev = reverse(day);
            if (Math.abs(day - rev) % k == 0) {
                beautiful_days++;
            }
        }
        System.out.println(beautiful_days);
    }
}

Beautiful Days at the Movies javaScript Solution

function reverse(n) {
    let rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n = Math.floor(n / 10);
    }
    return rev;
}

function beautifulDays(i, j, k) {
    let beautiful_days = 0;
    for (let day = i; day <= j; day++) {
        let rev = reverse(day);
        if (Math.abs(day - rev) % k === 0) {
            beautiful_days++;
        }
    }
    return beautiful_days;
}

console.log(beautifulDays(20, 23, 6)); // Output: 2

Beautiful Days at the Movies Python Solution

def reverse(n):
    rev = 0
    while n > 0:
        rev = rev * 10 + n % 10
        n //= 10
    return rev

def beautifulDays(i, j, k):
    beautiful_days = 0
    for day in range(i, j+1):
        rev = reverse(day)
        if abs(day - rev) % k == 0:
            beautiful_days += 1
    return beautiful_days

print(beautifulDays(20, 23, 6)) # Output: 2

Other Solution

c C# C++ HackerRank Solutions java javascript python

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