Example: Triplets with Given Sum - Brute Force in Java

// Triplets with Given Sum - Brute Force
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

// Main class containing the entry point of the program
public class Main {
    public static void main(String[] args) {
        int[] arr = {12, 3, 1, 2, -6, 5, -8, 6};
        int targetSum = 0;
        List<List<Integer>> result = findTripletsBruteForce(arr, targetSum);

        System.out.println("Triplets (Brute Force) for sum " + targetSum + ":");
        result.forEach(System.out::println);

        int[] arr2 = {1, 2, 3, 4, 5, 6};
        targetSum = 9;
        List<List<Integer>> result2 = findTripletsBruteForce(arr2, targetSum);
        System.out.println("\nTriplets (Brute Force) for sum " + targetSum + ":");
        result2.forEach(System.out::println);
    }

    public static List<List<Integer>> findTripletsBruteForce(int[] arr, int targetSum) {
        List<List<Integer>> triplets = new ArrayList<>();
        int n = arr.length;

        // Sort the array to handle duplicates later (optional for correctness but good for unique triplets)
        // Although sorting itself is O(N log N), the O(N^3) dominates.
        Arrays.sort(arr); 

        // Step 1: Iterate through the first element
        for (int i = 0; i < n - 2; i++) {
            // Skip duplicate elements for the first number
            if (i > 0 && arr[i] == arr[i - 1]) {
                continue;
            }

            // Step 2: Iterate through the second element
            for (int j = i + 1; j < n - 1; j++) {
                // Skip duplicate elements for the second number
                if (j > i + 1 && arr[j] == arr[j - 1]) {
                    continue;
                }

                // Step 3: Iterate through the third element
                for (int k = j + 1; k < n; k++) {
                    // Skip duplicate elements for the third number
                    if (k > j + 1 && arr[k] == arr[k - 1]) {
                        continue;
                    }

                    // Step 4: Check if the sum equals the target
                    if (arr[i] + arr[j] + arr[k] == targetSum) {
                        triplets.add(Arrays.asList(arr[i], arr[j], arr[k]));
                    }
                }
            }
        }
        return triplets;
    }
}