Below is the code I have written for counting perfect squares between a given lower and upper bound.I am using the following concept to solve this:Starting with 1, there are \$\sqrt{m}\$ square numbers up to and including \$m\$.public static int CountPerfectSquares(long A, long B){                int count = 0;    //Proceed if lowerbound is less than upperbound    if (A < B)    {        //negative numbers are not perfect squares        if (A < 0 && B < 0) return count;         //Reset lowerbound if required        if (A <= 0) A = 1;        //Find number of squares between A & B (INCLUSIVE lower and upperbound)        //count = (int)Math.Sqrt(B) - (int)Math.Sqrt(A);  //Giving wrong count when LowerBound value is a whole square (For exp. CountPerfectSquares(1, 25))        int sqrtA = (int)Math.Sqrt(A);        int sqrtB = (int)Math.Sqrt(B);        count = sqrtB - sqrtA;        //To handle the case when lowerbound value is a perfect square number        if (A == sqrtA * sqrtA)         {            count++;        }    }    return count;}Can you please help me make this more efficient, if possible?Is there any other way of handling the case when the lower bound value is a perfect square number, so that I don't have to do an extra count++?
Counting perfect squares
c#;mathematics
If A was not included in the observed range, you wouldn't need the if (A == sqrtA * sqrtA) part, right? So, let's use that: instead of checking the range {A,A+1,...,B}, check the range {A-1,A,...,B}. Since you're working with integers, the two are the same:        int sqrtA = (int)Math.Sqrt(A-1);        int sqrtB = (int)Math.Sqrt(B);        count = sqrtB - sqrtA;A minor detail: when returning a predefined constant, I prefer writing the constant itself, not a variable initialized to that value:        if (A < 0 && B < 0) return 0; It seems more readable to me.Also, I would replace two main if-s with if (B < 0 || B < A) return 0. This gives the same result, but in a cleaner way.So, my version would be:public static int CountPerfectSquares(long A, long B){                // Return zero if there are obviously no perfect squares    if (B < 0 || B < A) return 0;    // Reset the lower bound if required    if (A <= 0) A = 1;    // Find the number of perfect squares between A & B (INCLUSIVE)    return (int)Math.Sqrt(B) - (int)Math.Sqrt(A-1);  // Now giving the correct count, even when the lower bound value is a whole square (for exp. CountPerfectSquares(1, 25))}