The brute force approach will loop through all the way from 1 to N – looking for divisors and add them to a running sum.

public boolean checkPerfectNumber(int num) {
if (num <= 0) {
return false;
}
int sum = 0;
for (int i = 1; i * i <= num; i++) {
if (num % i == 0) {
sum += i;
if (i * i != num) {
sum += num / i;
}

}
}
return sum - num == num;
}
}

Only Loop till SqRt (N) – O(sqrt N)

There is no need to loop till N. Any divisor will need to be less than or equal to sqrt N. So – shorten the loop to only go till sqrt N.

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