15-3Sum.md

15. 3Sum - Solution Approach

Problem Overview🤔

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.

Solution Approach🎯

The solution uses a two-pointer technique combined with sorting to efficiently find triplets. Here's the step-by-step approach:

• Sort the input array first - this is crucial for avoiding duplicates and using the two-pointer technique effectively
• Iterate through the array with index i as the first number of the triplet
• For each i, use two pointers (left and right) to find pairs that sum to -nums[i]
• Skip duplicate values of i to avoid duplicate triplets

Key Points💡

• The array must be sorted first
• Duplicate handling is critical for correct results

Code Implementation💻

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

        for (int i = 0; i < n - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int left = i + 1, right = n - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0) {
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    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;
    }
}

Time and Space Complexity⚡

• Time Complexity: O(n²) where n is the length of input array
• Space Complexity: O(1) excluding the space required for output

Example Walkthrough

For input array: [-1, 0, 1, 2, -1, -4]

• After sorting: [-4, -1, -1, 0, 1, 2]
• First iteration: i = -4, looking for pairs that sum to 4
• Second iteration: i = -1 (first occurrence), looking for pairs that sum to 1
• Skip second -1 to avoid duplicates
• Continue with remaining elements

The solution will find all unique triplets: [[-1, -1, 2], [-1, 0, 1]]