Problem: Finding the number of perfect squares between two integer range.

Input: first integer and last integer -> int startInteger, int endInteger

Output: total number of perfect squares including this number -> int count

The worst approach would be to loop over each element and check whether it is an integer.

The best approach would be to determine the floor of root of first integer and ceil of root of last integer.

So lets take an example of 10, 100.

The expected answer would be: 7

So first compute the floor of the square root of 10 that would be 3.

And then compute the ceil of the square root of 100 that would be 10.

10 – 3 = 7 will be the output. It will be O(1) .

You can have below code from GIT.

**https://github.com/Niravkumar-Patel/SnippetExampleRepo.git**

import java.util.Scanner; public class PerfectSquare { public static void main(String args[]){ Scanner in = new Scanner(System.in); System.out.println("Please Enter Your Lower Limit"); int low = in.nextInt(); System.out.println("Please Enter Your Upper Limit"); int high = in.nextInt(); in.close(); int smallestNumber = (int)Math.ceil(Math.sqrt(low)); int highestNumber = (int)Math.floor(Math.sqrt(high)); System.out.println("Total Number of Perfect square are:"+(highestNumber-smallestNumber+1)); } }

The output of above code is as below:

Please Enter Your Lower Limit 10 Please Enter Your Upper Limit 100 Total Number of Perfect square are:7 Please Enter Your Lower Limit 2 Please Enter Your Upper Limit 12 Total Number of Perfect square are:2 Please Enter Your Lower Limit 45 Please Enter Your Upper Limit 500 Total Number of Perfect square are:16

You can have above code from GIT.