3Sum = 0 | Time: O(n^2) | Space: O(1)

/*  Leetcode problem: https://leetcode.com/problems/3sum/
    Used solution as reference: https://leetcode.com/problems/3sum/discuss/7399/Easiest-Java-Solution
*/

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class ThreeSum {

    public static List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> results = new ArrayList<>();
        Arrays.sort(nums);

        for (int i = 0; i < nums.length - 2; i++) {
            // skip same result
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }

            int left = i + 1;
            int right = nums.length - 1;
            int target = -nums[i];
            while(left < right) {
                if (nums[left] + nums[right] == target) {
                    results.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    left++;
                    right--;
                    while (left < right && nums[left] == nums [left -1]) left++; // skip the same result
                    while (left < right && nums[right] == nums [right + 1]) right--; // skip the same result
                } else if (nums[left] + nums [right] < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return results;
    }

    public static void main(String arg[]) {
        int[]testInput = {-1, 0, 1, 2, -1, -4};
        List<List<Integer>> result = threeSum(testInput);
        System.out.println("Results are: ");
        for (List<Integer> list : result) {
            System.out.println(list);
        }

    }
}