Square Root of a number

/*
    Time Complexity: O(log(n))
    Space Complexity: O(1)

    Where 'n' is the given number.
*/

// Function to find the square root of 'n'
int floorSqrt(int n){

    // Handle corner case
    if(n == 0 || n == 1)
    {
        return n;
    }

    // Initialize variable 'low' and 'high' with lower and upper bound of answer
    int low = 2;
    int high = n;

    // To store mid of 'low' and 'high'
    int mid;

    // Loop while 'low' is less than 'high'
    while(low < high)
    {
        // Calculating 'mid'
        mid = low + (high - low) / 2;

        /*
            If square of 'mid' is less than 'n' update 'low' else update 'high'
            To avoid overflow we have nOT used the condition as ('mid' * 'mid' <= n)
        */
        if(mid <= (n / mid))
        {
            low = mid + 1;
        }
        else
        {
            high = mid;
        }
    }

    // Return 'low' - 1
    return low - 1;
}