Given some numbers, we need to find whether they are smart numbers or not. If they are smart we should print “YES” otherwise print “NO”.

A smart number is defined as a number that has an odd number of factors.

We have to perform debugging of the code. We can only modify one line of code and can not add or delete any line.

## Given Code

```
import math
def is_smart_number(num):
val = int(math.sqrt(num))
if num / val == 1:
return True
return False
for _ in range(int(input())):
num = int(input())
ans = is_smart_number(num)
if ans:
print("YES")
else:
print("NO")
```

We can only modify one line in this code.

## Approach for Smart Number Hackerrank Solution

The different factors of a number form pairs among each other. If x is a factor of a number n, we know that ** n/x** will also be a factor of that number. For example, 2 is a factor of 10, then 10/2, 5 is also a factor of 10.

This is the approach that we use to find factors of a number in ** O(sqrt(n))** time complexity. We just loop through all numbers till

**. If it is a factor we add that number and also n divided by that number to the list of factors. So we always get a pair of factors.**

*sqrt(n)*The only time we can have an odd number of factors is if ** x = n/x**. Thus the factor is its own pair. This equation can be simplified as

**Or**

*n = x^2***.**

*x = sqrt(n)*So, if an integer square root of n exists that is if n is a perfect square, we get an odd number of factors, making it a special number.

Thus, we can update the condition to check for this.

* if num / val == 1: *

becomes

* if num / val == val:*

## Smart Number Hackerrank Solution

```
import math
def is_smart_number(num):
val = int(math.sqrt(num))
if num / val == val:
return True
return False
for _ in range(int(input())):
num = int(input())
ans = is_smart_number(num)
if ans:
print("YES")
else:
print("NO")
```

## Output:

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