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.

Leetcode Link

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.

Constraints:

• 3 <= nums.length <= 3000
• -105 <= nums[i] <= 105

Code

java Code

public  List<List<Integer>> threeSum(int[] nums) {
        Set<List<Integer>> res  = new HashSet<>();
        if(nums.length==0) return new ArrayList<>(res);
        Arrays.sort(nums);
        for(int i=0; i<nums.length-2;i++){
            int j =i+1;
           int  k = nums.length-1;
            while(j<k){
                int sum = nums[i]+nums[j]+nums[k];
                if(sum==0)res.add(Arrays.asList(nums[i],nums[j++],nums[k--]));
                else if (sum >0) k--;
                else if (sum<0) j++;
            }

        }
        return new ArrayList<>(res);

    }