3Sum-LeetCode

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

LeetCode Problem - 15

import java.util.*;

class Solution {
    // Method to find all unique triplets in the array 'nums' that sum up to zero
    public List<List<Integer>> threeSum(int[] nums) {
        // If array is null or has less than 3 elements, return an empty list
        if (nums == null || nums.length < 3)
            return new ArrayList<>();

        Arrays.sort(nums);  // Sort the array to efficiently find triplets
        Set<List<Integer>> answer = new HashSet<>();  // Set to store unique triplets

        // Iterate through the array to find triplets
        for (int i = 0; i < nums.length - 2; i++) {
            int left = i + 1;  // Pointer for the element after 'nums[i]'
            int right = nums.length - 1;  // Pointer for the last element in the array

            // Use two pointers technique to find pairs that sum up to -(nums[i])
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];  // Calculate the sum of triplet

                if (sum == 0) {
                    // If sum is zero, add the triplet to the answer set
                    answer.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    left++;  // Move the left pointer to the right to find next valid triplet
                    right--;  // Move the right pointer to the left to find next valid triplet
                } else if (sum < 0) {
                    // If sum is less than zero, move the left pointer to increase the sum
                    left++;
                } else {
                    // If sum is greater than zero, move the right pointer to decrease the sum
                    right--;
                }
            }
        }

        // Convert set of lists to a list of lists and return as the final answer
        return new ArrayList<>(answer);
    }
}