Day 26: Nested Logic | 30 Days Of Code | HackerRank Solution

Hello coders, today we are going to solve Day 26: Nested Logic HackerRank Solution in C++, Java and Python.

Objective

Today’s challenge puts your understanding of nested conditional statements to the test.

Task

Your local library needs your help! Given the expected and actual return dates for a library book, create a program that calculates the fine (if any). The fee structure is as follows:

1. If the book is returned on or before the expected return date, no fine will be charged (i.e.: fine = 0.
2. If the book is returned after the expected return day but still within the same calendar month and year as the expected return date, fine = 15 Hackos x (the number of days late).
3. If the book is returned after the expected return month but still within the same calendar year as the expected return date, the fine = 500 Hackos x (the number of months late).
4. If the book is returned after the calendar year in which it was expected, there is a fixed fine of 10000 Hackos.

Example
d1, m1, y1 = 12312014  returned date
d2, m2, y2 = 112015 due date

The book is returned on time, so no fine is applied.

d1, m1, y1 = 112015 returned date
d2, m2, y2 = 12312014 due date

The book is returned in the following year, so the fine is a fixed 10000.

Input Format

The first line contains 3 space-separated integers denoting the respective day, month, and year on which the book was actually returned.
The second line contains 3 space-separated integers denoting the respective day, month, and year on which the book was expected to be returned (due date).

Constraints

• 1 <= D <= 31
• 1 <= M <= 12
• 1 <= Y <= 3000
• It is guaranteed that the dates will be valid Gregorian calendar dates.

Output Format

Print a single integer denoting the library fine for the book received as input.

Sample Input

STDIN       Function
-----       --------
9 6 2015    day = 9, month = 6, year = 2015 (date returned)
6 6 2015    day = 6, month = 6, year = 2015 (date due)

Sample Output

45

Explanation

Given the following return dates:
Returned: D₁ = 9, M₁ = 6, Y₁ = 2015
Due: D₂ = 6, M₂ = 6, Y₂ = 2015

Because Y₂ = Y₁, it is less than a year late.
Because M₂ = M₁, it is less than a month late.
Because D₂ < D₁, it was returned late (but still within the same month and year).

Per the library’s fee structure, we know that our fine will be 15 Hackos x (#dayslate). We then print the result of 15 x (D₁ – D₂) = 15 x (9 – 6) = 45 as our output.

Solution – Day 26: Nested Logic

C++

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

int main() 
{
    int Day, Month, Year, Day1, Month1, Year1, mdiff, ddiff, ydiff, diff;
    
    cin>>Day>>Month>>Year>>Day1>>Month1>>Year1;

    mdiff = Month - Month1;
    ddiff = Day - Day1;
    ydiff = Year - Year1;
    diff=(Year-Year1>0)?10000:(Month-Month1>0&&ydiff==0)?mdiff*500:(Day-Day1>0 && ydiff==0)?ddiff*15:0;    
    cout<<diff<<endl;

    return 0;
}

Java

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int day = sc.nextInt();
        int month = sc.nextInt();
        int year = sc.nextInt();
        int dayExpected = sc.nextInt();
        int monthExpected = sc.nextInt();
        int yearExpected = sc.nextInt();
        int fine = 0;
        
        if(year > yearExpected){
            fine = 10000;
        }else if(year == yearExpected && month > monthExpected){
            fine = (month - monthExpected) * 500;
        }else if(year == yearExpected && month == monthExpected && day > dayExpected){
            fine = (day - dayExpected) * 15;
        }
        System.out.println(fine);
    }

}

Python

actual = input()
actual = list(map(int, actual.split(" ")))

expected = input()
expected = list(map(int, expected.split(" ")))

actualDate = actual[0]
actualMonth = actual[1]
actualYear = actual[2]

expectedDate = expected[0]
expectedMonth = expected[1]
expectedYear = expected[2]

fine = 0

if actualYear > expectedYear:
    fine = 10000
elif actualYear == expectedYear: #Same calender year
    if actualMonth > expectedMonth: #checking months
        fine = 500 * (actualMonth - expectedMonth)
    elif actualMonth == expectedMonth: #check dates now
        if actualDate > expectedDate:
            fine = 15 * (actualDate - expectedDate)
print(fine)

Disclaimer: The above Problem (Day 26: Nested Logic) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.