LeetCode Study Note | Problem 15: 3Sum

✨ Problem Description

Given an integer array nums, return all the unique triplets [nums[i], nums[j], nums[k]] such that i ≠ j ≠ k, and nums[i] + nums[j] + nums[k] == 0.

🔍 Constraints:

• Solution set must not contain duplicate triplets.

• The answer should be a list of triplets, not index-based.

• Input array can have both positive and negative numbers.

🧩 Core Idea

Although this problem sounds like a straightforward extension of the classic Two Sum, it introduces new challenges:

• Need to find all combinations, not just one.

• Must eliminate duplicates.

• Return triplets of values, not indices.

This changes the optimal strategy from hash-based lookup to a sort + two-pointer approach.

🧪 Why Not Use HashMap like Two Sum?

In Two Sum:

• Only one valid pair is needed.

• Index-based lookup is prioritized.

• No duplicates or ordering constraints.

In 3Sum:

• Need all unique value triplets.

• Sorting makes it easier to skip duplicates.

• Two-pointer technique becomes efficient after sorting.

So in short:

✅ Two Sum → HashMap is optimal

✅ Three Sum → Sorting + Two Pointers is optimal

✅ Java Solution: 3Sum with Sorting + Two Pointers

import java.util.*;

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums); // Step 1: sort the array

        for (int i = 0; i < nums.length - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue; // Skip duplicate a

            int target = -nums[i];
            int left = i + 1, right = nums.length - 1;

            while (left < right) {
                int sum = nums[left] + nums[right];
                if (sum == target) {
                    res.add(Arrays.asList(nums[i], nums[left], nums[right]));

                    // Skip duplicates
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    while (left < right && nums[right] == nums[right - 1]) right--;

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

        return res;
    }
}

📎 Related Concept: Two Sum with Unique Pairs (Extended Insight)

What if Two Sum also needed:

• All pairs (not just one)?

• No duplicates?

• Only value-based results?

Then it would require sorting + two pointers, just like 3Sum.

class Solution {
    public List<List<Integer>> twoSumUniquePairs(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);

        int left = 0, right = nums.length - 1;

        while (left < right) {
            int sum = nums[left] + nums[right];
            if (sum == target) {
                res.add(Arrays.asList(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 < target) {
                left++;
            } else {
                right--;
            }
        }

        return res;
    }
}

📌 Key Takeaways