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How you can Discover the Squares in Java


The problem

Full the perform that takes an odd integer (0 < n < 1000000) which is the distinction between two consecutive excellent squares, and return these squares as a string within the formatĀ "bigger-smaller".

Examples

9  -->  "25-16"
5  -->  "9-4"
7  -->  "16-9"

The answer in Java code

Possibility 1:

public class Resolution {
  public static String findSquares(closing int n) {
    closing lengthy a = (n + 1) / 2;
    closing lengthy b = a - 1;
    return String.format("%d-%d", a * a, b * b);
  }
}

Possibility 2:

interface Resolution {
  static String findSquares(lengthy n) {
    return (n /= 2) * n + 2 * n + 1 + "-" + n * n;
  }
}

Possibility 3:

public class Resolution{
  public static String findSquares(int n){
    lengthy n1 = n/2;
    lengthy n2 = n1+1;
    return String.valueOf(n2*n2)+"-"+ String.valueOf(n1*n1);
  }
}

Take a look at circumstances to validate our resolution

import org.junit.Take a look at;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;

public class SolutionTest {
    @Take a look at
    public void testBasicNumbers() {
        assertEquals("25-16",Resolution.findSquares(9));
    }
    @Take a look at
    public void testSmallerNumbers() {
        assertEquals("1-0",Resolution.findSquares(1));
    }
    @Take a look at
    public void testBiggerNumbers() {
        assertEquals("891136-889249",Resolution.findSquares(1887));
        assertEquals("2499600016-2499500025",Resolution.findSquares(99991));
    }
}

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