15. 3Sum

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) {
        // Assumptions: array is not null, array.length >= 3

        List<List<Integer>> result = new ArrayList<List<Integer>>();

        Arrays.sort(nums);
        int target = 0;
        for (int i = 0; i < nums.length -2; i++){
            //The goal is to find i < j < k, such that
            // nums[i] + nums[j] + nums[k] == 0;
            // To make sure there is no duplicate possible i.
            // e.g. if we have 2,2,2 only the first 2 will be selected as i.
            if (i > 0 && nums[i] == nums[i-1]){
                continue;
            }

            int left = i + 1;
            int right = nums.length - 1;
            while(left < right){
                int tmp = nums[left] + nums[right];
                if (nums[i] + tmp == target){
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    left++;
                    // ignore all possibel duplicate j as well
                    while (left < right && nums[left] == nums[left -1]){
                        left++;
                    }
                }else if (tmp + nums[i] < target){
                    left++;
                }else {
                    right--;
                }

            }
        }
        return result;

    }
}

// TC: O(n^2)
// SC: O(n)

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();

        // Sort the array
        Arrays.sort(nums);

        for (int i = 0; i < nums.length -2; i++){
            // Skip duplicate elements for i

            if (i > 0 && nums[i] == nums[i-1]){
                continue;
            }

            int j = i + 1;
            int k = nums.length - 1;

            while (j < k){
                int sum = nums[i] + nums[j] + nums[k];

                if (sum == 0){
                    result.add(Arrays.asList(nums[i], nums[j], nums[k]));

                    // Skip duplicate elements for j
                    while(j <k && nums[j] == nums[j+1]){
                        j++;
                    }
                    // Move the pointers
                    j++;
                    k--;
                }else if (sum < 0){
                    // Sum is less than zero, increment j to increase the sum
                    j++;
                }else {
                    // Sum is greater than zero, decrement k to decrease the sum
                    k--;
                }
            }
        }
        return result;
    }
}

// TC: O(n^2)

// SC: O(n) // sort quick sort O(n) 

/*

 [-1, -1, 2], [-1, 0, 1]
    [-1,0,1,2,-1,-4]

    [-4, -1, -1, 0 , 1 , 2]

                 i 
                    l            
                    r

*/