Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Output: true

Example 2:

Input: 14
Output: false

 class Solution {
    public boolean isPerfectSquare(int num) {
       for (long i = 1; i * i <= num; i++) {
  
            if ((i)*(i) == num) {
                return true;
            }
        }
        return false;
       
    }
}

Try it on Leetcode

Here, they mentioned in problem not to use in-build methods. We are going to use basic idea to solve this problem. We are going to loop i from 1 to num. If i*i == num return true, else iterate till the full loop.

Below the points to be noted:

Variable used for iterating for loop must be of type long , because if to avoid overflow in int type. Also no need to iterate till num and a condition should be i*i<=num [i.e., if i*i is greater than num no need of checking for perfect square and we can exit from loop]

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