square numbers Java

Question

"An array is used to store ten integer numbers. Write a Java program that determines and print the square numbers which are also odd numbers from the given array."

The problem I have is how to figure out if the number in the array is a square number. I tried this way but it's not correct!

import java.math.*;
public class JavaApplication43 {

    public static void main(String[] args) {

        int[] no = {22, 44, 25, 89, 81, 55, 23, 25, 55};

        for (int i = 0; i < no.length; i++) {

            int x = no[i];
            double y;
            if (x % 2 != 0) {
                y = Math.sqrt(x);
                if (x == (Math.pow(y, 2))) 
                    System.out.println(no[i]);
            }
       }
   }
}

This is the output it gives me

run:
25
81
55
25
55

55 is there too that means this method I used is not successful!

I guess you mean the numbers must be squares of integers only since every positive number is a square of some real number. Hence you could check whether `y` is an integer (i.e. the fraction part is 0 or at least very small because of precision issues). — Thomas, Apr 18 '16 at 08:20
Check this [link](http://stackoverflow.com/questions/295579/fastest-way-to-determine-if-an-integers-square-root-is-an-integer) — greenPadawan, Apr 18 '16 at 08:23

Konstantin Yovkov · Accepted Answer · 2016-04-18T08:21:36.670

5

You could just do:

for (int i = 0; i < no.length; i++) {
    int x = no[i];
    if (x % 2 == 0) continue;
    int y = (int) Math.sqrt(x);
    if (x == y * y) { 
        System.out.println(x);
    }
}

edited Apr 18 '16 at 08:21

answered Apr 18 '16 at 08:20

Konstantin Yovkov

62,134
8
100
147

Why shouldn't he eliminate even numbers if that's a requirement? – Thomas Apr 18 '16 at 08:21
@PeterLawrey, for example? :) – Konstantin Yovkov Apr 18 '16 at 08:22
1

@KonstantinYovkov the number would have to be large, larger than you can get in an `int` value come to think of it. ;) – Peter Lawrey Apr 18 '16 at 08:23
but using integers in the array – isharailanga Apr 18 '16 at 08:23
@KonstantinYovkov due to precision issues you could get a fraction of `.99999999` instead of the next higher integer and trunction would thus result in the wrong value. – Thomas Apr 18 '16 at 08:24
1

@laish129 the square root of an integer in most cases is still a double though. – Thomas Apr 18 '16 at 08:24

Peter Lawrey · Answer 2 · 2016-04-18T08:27:17.303

5

You can determine if a number is a square by checking if its square root is an integer.

double sqrt = Math.sqrt(x);
long sqrt2 = Math.round(sqrt);
if (Math.abs(sqrt - sqrt2) / sqrt < 1e-15)
   // we have a square.

If you know x is an int you shouldn't get a rounding error and you can do

int x = ...
double sqrt = Math.sqrt(x);
if ((int) sqrt == sqrt)
    // we have a square.

edited Apr 18 '16 at 08:27

answered Apr 18 '16 at 08:21

Peter Lawrey

525,659
79
751
1,130

This pattern can be used to the limits of `double` precision although `int` won't test those limits. – Peter Lawrey Apr 18 '16 at 08:26

Isuru · Answer 3 · 2016-04-18T16:31:24.097

hope this will be much easier,

    if((arr[i]%2 != 0) & (Math.sqrt(arr[i])%1 == 0)){
        System.out.println(arr[i]);
    }

In here inside if condition first I check whether number is an odd number by taking modulus division and the 2nd condition check whether number is a perfect square. First I get the square root of given number and then take the modulus division by 1 and check whether it's equal to 0. If number is a perfect square, square root of that number is an integer, when I take the modulus division of an integer answer should be equal to 0.

square numbers Java

3 Answers3