How to Get the N-th Power in Java

The challenge

You are given an array with positive numbers and a number N. You should find the N-th power of the element in the array with the index N. If N is outside of the array, then return -1. Don’t forget that the first element has the index 0.

Let’s look at a few examples:

  • array = [1, 2, 3, 4] and N = 2, then the result is 3^2 == 9;
  • array = [1, 2, 3] and N = 3, but N is outside of the array, so the result is -1.

The solution in Java code

public class Solution { public static int nthPower(int[] array, int n) { return n >= array.length ? -1 : (int) Math.pow(array[n], n); } }
Code language: Java (java)

An alternative mapping:

public class Solution { public static int nthPower(int[] array, int n) { if (n < 0 || n >= array.length) return -1; return (int) Math.pow(array[n], n); } }
Code language: Java (java)

Test cases to validate our Java code

import org.junit.Test; import static org.junit.Assert.assertEquals; import org.junit.runners.JUnit4; import java.util.stream.IntStream; public class SolutionTest { @Test public void basicTests() { assertEquals(-1, Solution.nthPower(new int[] {1,2}, 2)); assertEquals(8, Solution.nthPower(new int[] {3,1,2,2}, 3)); assertEquals(4, Solution.nthPower(new int[] {3,1,2}, 2)); } @Test public void advancedTests() { assertEquals(9261, Solution.nthPower(new int[] {7,14,9,21,15}, 3)); assertEquals(169, Solution.nthPower(new int[] {2,7,13,17}, 2)); assertEquals(50625, Solution.nthPower(new int[] {11,23,3,4,15,112,12,4}, 4)); } @Test public void moreTests() { assertEquals(1, Solution.nthPower(new int[] {2,1,2,1,2,1}, 5)); assertEquals(-1, Solution.nthPower(new int[] {11,23,3,4,15}, 7)); assertEquals(-1, Solution.nthPower(new int[] {3,2,1,2,3,1}, 6)); } public int nthPowerSol(int[] array, int n) { return n >= array.length ? -1 : (int) Math.pow(array[n], n); } private static int randomInRange(int min, int max) { int range = (max - min) + 1; return (int)(Math.random() * range) + min; } @Test public void randomTests() { int[] array = new int[randomInRange(2,5)]; int n = randomInRange(2,6); for (int i = 0; i < 50; i++) { for (int k = 0; k < array.length; k++) { array[k] = randomInRange(2,15); } assertEquals(nthPowerSol(array, n), Solution.nthPower(array, n)); } } }
Code language: Java (java)
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