3Sum medium

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: Because there are two triplets whose sum is 0: (-1) + (-1) + 2 = 0 (-1) + 0 + 1 = 0

Example 2

Input: nums = [0,1,1] Output: [] Explanation: There are no triplets whose sum is 0

Steps

1. Sort the array: Sorting allows us to easily skip duplicate numbers and optimize the search for triplets.

2. Iterate through the array: The outer loop iterates through each number nums[i] as a potential first element of a triplet.

3. Two-pointer approach: For each nums[i], we use two pointers, left and right, to search for the remaining two numbers that sum to -nums[i]. left starts at i + 1 and right starts at the end of the array.

4. Sum and adjust pointers: We calculate the sum sum = nums[i] + nums[left] + nums[right].

   • If sum is 0, we've found a triplet. Add it to the result set, but make sure to skip duplicate numbers by incrementing left and decrementing right until we encounter unique numbers.
   • If sum is less than 0, it means we need a larger sum, so we increment left.
   • If sum is greater than 0, it means we need a smaller sum, so we decrement right.

5. Skip duplicates: After finding a triplet, we increment left and decrement right while they point to duplicate numbers to avoid adding duplicate triplets to the result.

Explanation

The algorithm uses a combination of sorting and a two-pointer technique to efficiently find all unique triplets that sum to zero. Sorting enables easy duplicate handling and efficient searching using the two pointers. The two-pointer approach allows for a linear time complexity within the nested loop, resulting in an overall time complexity of O(n^2).

Code

using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
    public IList<IList<int>> ThreeSum(int[] nums) {
        Array.Sort(nums); // Sort the array
        IList<IList<int>> result = new List<IList<int>>();

        for (int i = 0; i < nums.Length - 2; i++) {
            // Skip duplicate numbers for the first element
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            int left = i + 1;
            int right = nums.Length - 1;

            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];

                if (sum == 0) {
                    result.Add(new List<int> { nums[i], nums[left], nums[right] });

                    // Skip duplicate numbers for the second and third elements
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    while (left < right && nums[right] == nums[right - 1]) right--;

                    left++;
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }

        return result;
    }
}

Complexity

• Time Complexity: O(n^2), where n is the length of the input array. The sorting takes O(n log n), but the nested loops dominate the time complexity.
• Space Complexity: O(log n) in the best case (due to the space used by the sorting algorithm, which can be in-place for some implementations), or O(n) in the worst case for certain sorting algorithms, and O(m) where m is the number of triplets found. The space used by the result list depends on the number of triplets. In most scenarios, m will be significantly smaller than n.