# (0) Binary Search: Valid Perfect Square

 [Question Link](https://leetcode.com/problems/valid-perfect-square/) and  [Solution Link](https://leetcode.com/submissions/detail/610152800/) 

**Difficulty**: Easy

**Problem Statement**: Given positive number return true if the number is perfect square else return false.

**Approach1**: Use a simple while loop and traverse through all numbers starting from 1 (cur) and check if `cur*cur == num`. If at any point it's true then return true else after the loop returns false.

```
bool isPerfectSquare(int num)
{
    long long cur = 1;
    while (cur * cur <= num)
    {
        if (cur * cur == num) return true;
        cur++;
    }
    return false;
}
``` 

**Approach2**: Use binary search. Make `low=1` and `high=2^31-1` because our lowest value can be 1 and the highest can be `2^31-1`. Now every time calculate mid. If `mid*mid` is equal to num then return true. Else if `mid*mid` is greater than num then `high=mid-1` else `low=mid+1`.

```
bool isPerfectSquare(int num)
{
    long long low = 1, high = (1 << 31 - 1);
    while (low <= high)
    {
        long long mid = low + (high - low) / 2;
        long long ans = mid * mid;
        if (ans == num) return true;
        else if (ans > num) high = mid - 1;
        else low = mid + 1;
    }
    return false;
}
```