3. 3Sum

Description

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.

Constraints:

• 3 <= nums.length <= 3000
• -10^5 <= nums[i] <= 10^5

Approach

• Sort the array and start first loop then perform 2 Sum of sorted array let’s say a, b, c
• For Optimization if a is positive then sum = 0 cannot be achieved as this is sorted now
• Next Optimization for duplicate is you check if current equals current-1 then move pointer to right
• Using LinkedList could be faster as it takes O(1) for add and remove element as compared to ArrayList O(n)
• Time: O(n^2) Space: O(1)
• Sorting takes O(nlogn) but since it is loop inside loop so O(n^2) triumphs
• Space complexity could be O(n) depending on sorting logic

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> ans = new LinkedList<List<Integer>>();
        Arrays.sort(nums);
 
        for (int i = 0; i < nums.length - 2; i++) {
            int a = nums[i];
 
            if(a > 0) //sorted array so no more solutions
                continue; //can be replaced with break;
 
            if (i > 0 && a == nums[i-1])
                continue; //skip duplicates
 
            int l = i + 1, r = nums.length - 1;
            while (l < r) {
                int sum = a + nums[l] + nums[r];
                if (sum < 0)
                    l++;
                else if (sum > 0)
                    r--;
                else {
                    ans.add(List.of(a, nums[l], nums[r]));
                    l++;
                    r--;
                    while(nums[l] == nums[l-1] && l < r)
                        l++; //skip duplicates
                }        
            }
        }
 
        return ans;
    }
}