[LeetCode] 15. 3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note: The solution set must not contain duplicate triplets.
For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Thought process:
Sort the array. Iterate through the sorted array. For each element, iterate through the elements to its right using two pointers: one from the right of this element, the other from end of the array.
  1. If nums[i] + nums[j] + nums[k] < 0, increment the left pointer.
  2. If equal, add indices to result and update both pointers.
  3. Otherwise, decrement the right pointer.
To avoid duplicate triplets, increment every pointer until the number is different from the previous one.

Solution:
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class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> triplets = new ArrayList<>();
        
        for (int i = 0; i < nums.length; i++) {
            while (i > 0 && i < nums.length && nums[i] == nums[i - 1]) {
                i++;
            }
            
            int j = i + 1;
            int k = nums.length - 1;
            while (j < k) {
                while (j > i + 1 && j < k && nums[j] == nums[j - 1]) {
                    j++;
                }
                while (k < nums.length - 1 && k > j && nums[k] == nums[k + 1]) {
                    k--;
                }
                if (j == k) {
                    break;
                }
                
                int sum = nums[i] + nums[j] + nums[k];
                if (sum < 0) {
                    j++;
                } else if (sum == 0) {
                    List<Integer> triplet = new ArrayList<>();
                    triplet.add(nums[i]);
                    triplet.add(nums[j]);
                    triplet.add(nums[k]);
                    triplets.add(triplet);
                    j++;
                    k--;
                } else {
                    k--;
                }
            }
        }
        return triplets;
    }
}

Time complexity:
Sorting takes O(nlogn). Loop takes O(n^2). The overall time complexity is O(n^2).
Location: San Jose, CA, USA

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