How to Find the Area or Perimeter of a 4-sided Polygon using Java

The challenge

You are given the length and width of a 4-sided polygon. The polygon can either be a rectangle or a square.
If it is a square, return its area. If it is a rectangle, return its perimeter.

area_or_perimeter(6, 10) --> 32 area_or_perimeter(4, 4) --> 16

Note: for the purposes of this challenge you will assume that it is a square if its length and width are equal, otherwise, it is a rectangle.

The solution in Java

Given that we know that:

The area of a square is: l x h and the area of a rectangle is 2(l+h), we can create the following code, differentiating on the if conditional statement that l==w, or, length is equal to width, which would mean a square, otherwise a rectangle:

public class Solution { public static int areaOrPerimeter (int l, int w) { if (l==w) { //square return l*w; } else { //rect return (l+w)*2; } } }
Code language: Java (java)

We can simplify this as follows:

public class Solution { public static int areaOrPerimeter (int l, int w) { return (l==w) ? l*w : (l+w)*2; } }
Code language: Java (java)

Test cases to validate our code

import org.junit.Test; import static org.junit.Assert.assertEquals; import org.junit.runners.JUnit4; public class SolutionTest { private static int X(int a , int b) { return a == b ? a * b : 2 * ( a + b ); } @Test public void Tests() { assertEquals(16, Solution.areaOrPerimeter(4 , 4)); assertEquals(32, Solution.areaOrPerimeter(6, 10), 32); for(int i = 1; i < 101; i++) { int a = (int)(Math. random() * (i + 50) + 1); int b = (int)(Math. random() * (i + 30) + 21); System.out.println("Testing for " + a + ", " + b); assertEquals(X(a, b), Solution.areaOrPerimeter(a, b)); } } }
Code language: Java (java)
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