# Smart Number 2 HackerRank Solution Smart Number 2 HackerRank Solution

In this Smart Number 2 HackerRank solution, our challenge 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.

Note: You can modify only one line in the given code and you cannot add or remove any new lines.

To restore the original code, click on the icon to the right of the language selector.

Input Format

The first line of the input contains , the number of test cases.
The next  lines contain one integer each.

Constraints

• , where  is the  integer.

Output Format

The output should consist of  lines. In the  line print YES if the  integer has an odd number of factors, else print NO.

Sample Input

4
1
2
7
169


Sample Output

YES
NO
NO
YES


Explanation

The factors of 1 are just 1 itself.So the answer is YES. The factors of 2 are 1 and 2.It has even number of factors.The answer is NO. The factors of 7 are 1 and 7.It has even number of factors.The answer is NO. The factors of 169 are 1,13 and 169.It has odd number of factors.The answer is YES.

## Smart Number 2 HackerRank Solution

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

### Problem Solution in Python

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

### Problem Solution in JavaScript

function processData(input) {
const arr = input.split('\n').slice(1)
arr.forEach(num => {
let count = 0
parsed = parseInt(num)
for(i=1; i <= parsed; i++){

if(parsed % i === 0) {
count ++
}
}
if(count % 2 === 0){
console.log("NO")
} else {
console.log("YES")
}
count = 0
} )
}
Note: You will never be able to pass this challenge unless the language you select already has the is_smart_number function in it already, but here is a JS solution.

### Problem Solution in Java

 public static boolean isSmartNumber(int num) {
int val = (int) Math.sqrt(num);
if((float)num / val == val)
return true;
return false;
}
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