Example: Maximum Product Subarray - Brute Force in Java

// Maximum Product Subarray - Brute Force
import java.util.Arrays;

// Main class containing the entry point of the program
public class Main {
    public static void main(String[] args) {
        int[] nums1 = {2, 3, -2, 4};
        System.out.println("Array: " + Arrays.toString(nums1) + ", Max Product: " + maxProductBruteForce(nums1)); // Expected: 6

        int[] nums2 = {-2, 0, -1};
        System.out.println("Array: " + Arrays.toString(nums2) + ", Max Product: " + maxProductBruteForce(nums2)); // Expected: 0

        int[] nums3 = {-2, 3, -4};
        System.out.println("Array: " + Arrays.toString(nums3) + ", Max Product: " + maxProductBruteForce(nums3)); // Expected: 24
    }

    public static int maxProductBruteForce(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0; // Or throw an exception, depending on requirements
        }

        int maxProduct = nums[0]; // Initialize with the first element

        // Step 1: Iterate through all possible starting points (i)
        for (int i = 0; i < nums.length; i++) {
            int currentProduct = 1; // Initialize current product for each new subarray

            // Step 2: Iterate through all possible ending points (j) from the current start
            for (int j = i; j < nums.length; j++) {
                // Step 3: Multiply the current element to the current product
                currentProduct *= nums[j];

                // Step 4: Update maxProduct if currentProduct is greater
                if (currentProduct > maxProduct) {
                    maxProduct = currentProduct;
                }
            }
        }
        return maxProduct;
    }
}