HackerRank Smart Number 2 problem solution

In this HackerRank Smart Number 2 problem solution, the task is to debug the existing code to successfully execute all provided test files.

A number is called a smart number if it has an odd number of factors. Given some numbers, find whether they are smart numbers or not.

Debug the given function is_smart_number to correctly check if a given number is a smart number.

Problem solution in Python.

import math

def is_smart_number(num):
    val = int(math.sqrt(num))
    if val ** 2 == num:
        return True
    return False

for _ in range(int(input())):
    num = int(input())
    ans = is_smart_number(num)
    if ans:
        print("YES")
    else:
        print("NO")

Problem solution in Java.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    public static boolean isSmartNumber(int num) {
        int val = (int) Math.sqrt(num);    
        if(Math.pow(val, 2) == num)
            return true;
        return false;
    }
    
    public static void main(String[] args) {
        int test_cases;
        Scanner in = new Scanner(System.in);
        test_cases = in.nextInt();
        int num;
        for(int i = 0; i < test_cases; i++){
            num = in.nextInt();
            boolean ans = isSmartNumber(num);
            if(ans){
                System.out.println("YES");
            }
            else System.out.println("NO");
        }
    }
}

Problem solution in C++.

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

bool is_smart_number(int num) {
    int val = (int) sqrt(num);
    if(val - (int)sqrt(num-1)!= 0)
        return true;
    return false;
}

int main() {
    
    int test_cases;
    cin >> test_cases;
    int num;
    for(int i = 0; i < test_cases; i++) {
        cin >> num;
        bool ans = is_smart_number(num);
        if(ans) {
            cout << "YES" << endl;
        }
        else cout << "NO" << endl;
    }
    return 0;
}