3 Sum

Problem Statement:

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.

Example 1:

Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0] Output: 0,0,0 Explanation: The only possible triplet sums up to 0.

Solution:

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        
        for(int i = 0; i < nums.length; i++){
            int target = -nums[i];
            int left = i + 1;
            int right = nums.length - 1;
 
            //if same target is achieved already, don't try it again.
            if(i > 0 && nums[i-1] == nums[i])
                continue;
 
            while(left < right){
                if(nums[left] + nums[right] < target){
                    left++;
                } else if(nums[left] + nums[right] > target){
                    right--;
                } else{
                    List<Integer> triplet = new ArrayList<>(Arrays.asList(nums[i], nums[left], nums[right]));
                    res.add(triplet);
 
                    //To avoid duplicate left and right values for the same target, 
                    //move 'em different numbers to skip duplicates
                    while(left < right && nums[right] == nums[right - 1]) right--;
                    while(left < right && nums[left] == nums[left + 1]) left++;
 
                    //Move left and right to valid triplets
                    left++; right--;
                }
            }
        }
 
        return res;
    }
}