Day 17: More Exceptions | 30 Days Of Code | HackerRank Solution

Hello coders, today we are going to solve Day 17: More Exceptions HackerRank Solution in C++, Java and Python.

Objective

Yesterday’s challenge taught you to manage exceptional situations by using try and catch blocks. In today’s challenge, you will practice throwing and propagating an exception.

Task

Write a Calculator class with a single method: int power(int,int). The power method takes two integers, n and p, as parameters and returns the integer result of n^p. If either n or p is negative, then the method must throw an exception with the message: n and p should be non-negative.

Note: Do not use an access modifier (e.g.: public) in the declaration for your Calculator class.

Input Format

Input from stdin is handled for you by the locked stub code in your editor. The first line contains an integer, , the number of test cases. Each of the T subsequent lines describes a test case in 2 space-separated integers that denote n and p, respectively.

Constraints

• No Test Case will result in overflow for correctly written code.

Output Format

Output to stdout is handled for you by the locked stub code in your editor. There are T lines of output, where each line contains the result of n^p as calculated by your Calculator class’ power method.

Sample Input

4
3 5
2 4
-1 -2
-1 3

Sample Output

243
16
n and p should be non-negative
n and p should be non-negative

Explanation

T = 4
T₀: 3 and 5 are positive, so power returns the result of 3⁵, which is 243.
T₁: 2 and 4 are positive, so power returns the result of 2⁴ =, which is 16.
T₂: Both inputs (-1 and -2) are negative, so power throws an exception and n and p should be non-negative is printed.
T₃: One of the inputs (-1) is negative, so power throws an exception and n and p should be non-negative is printed.

Solution – Day 17: More Exceptions

C++

#include <cmath>
#include <iostream>
#include <exception>
#include <stdexcept>
using namespace std;

//Write your code here
class Calculator 
{
    public:
    int power(int first, int second) 
    {
        if(first < 0 || second < 0) 
        {
           throw runtime_error("n and p should be non-negative");
        }
        return pow(first, second);
    }
};
int main()
{
    Calculator myCalculator=Calculator();
    int T,n,p;
    cin>>T;
    while(T-->0){
      if(scanf("%d %d",&n,&p)==2){
         try{
               int ans=myCalculator.power(n,p);
               cout<<ans<<endl; 
         }
         catch(exception& e){
             cout<<e.what()<<endl;
         }
      }
    }
    
}

Java

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

class Calculator {
    public int power(int n,int p) throws Exception{
        if(n<0 || p<0){
            throw new Exception("n and p should be non-negative");
        }
        else{
            return (int)Math.pow(n,p);
        }
    }
}
class Solution{
    public static void main(String[] args) {
    
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) {
        
            int n = in.nextInt();
            int p = in.nextInt();
            Calculator myCalculator = new Calculator();
            try {
                int ans = myCalculator.power(n, p);
                System.out.println(ans);
            }
            catch (Exception e) {
                System.out.println(e.getMessage());
            }
        }
        in.close();
    }

}

Python

class Calculator:
    def power(self, n, p):
        if n < 0 or p < 0:
            raise Exception("n and p should be non-negative")
        return pow(n, p)
myCalculator=Calculator()
T=int(input())
for i in range(T):
    n,p = map(int, input().split())
    try:
        ans=myCalculator.power(n,p)
        print(ans)
    except Exception as e:
        print(e)   

Disclaimer: The above Problem (Day 17: More Exceptions) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.