3Sum: Sorting Plus Two Pointers, With Duplicate Handling

"3Sum" shows up again and again in Meta, Amazon, Google phone screens. It is a Medium problem on paper, but the real test is whether you recognize the Two Pointers pattern quickly and code it cleanly. Part 7 of 12 in the Two Pointers arc.

The Problem

Given an integer array nums, return all unique triplets that sum to zero. The solution set must not contain duplicate triplets — and the deduplication, not the search, is where candidates lose this question.

Input:  nums = [-1, 0, 1, 2, -1, -4]
Output: [[-1, -1, 2], [-1, 0, 1]]

Recognizing the Two Pointers Pattern

Two pointers means walking a sequence with two indices that move based on a comparison — from both ends inward, or slow-and-fast in one direction. It turns brute-force pair scans that cost O(n²) into a single O(n) pass, and it is usually the intended answer whenever the input is sorted or can be sorted cheaply.

A k-sum problem reduces to (k−1)-sum with an outer loop: fix one number, then find pairs summing to its negation. After sorting, that inner pair-search is exactly the sorted Two Sum from the start of this arc.

The Approach

Sort. For each anchor index i (skipping repeated anchor values), run converging pointers on the suffix looking for -nums[i]. On a hit, record the triplet, then advance both pointers past any repeated values to avoid duplicate output.

Two pruning wins worth saying: a positive anchor ends the search (everything after is larger), and duplicate anchors are skipped with one comparison against the previous value.

Python Solution

def three_sum(nums: list[int]) -> list[list[int]]:
    """All unique triplets summing to zero."""
    nums.sort()
    n = len(nums)
    result = []
    for i in range(n - 2):
        if nums[i] > 0:
            break                      # sorted: no zero-sum possible
        if i > 0 and nums[i] == nums[i - 1]:
            continue                   # duplicate anchor
        left, right = i + 1, n - 1
        target = -nums[i]
        while left < right:
            s = nums[left] + nums[right]
            if s == target:
                result.append([nums[i], nums[left], nums[right]])
                left += 1
                right -= 1
                while left < right and nums[left] == nums[left - 1]:
                    left += 1
                while left < right and nums[right] == nums[right + 1]:
                    right -= 1
            elif s < target:
                left += 1
            else:
                right -= 1
    return result

Complexity

• Time: O(n²) — n anchors, each with a linear pointer sweep; the sort is dominated
• Space: O(1) — in-place sort and pointers, excluding the output list

Interview Tips and Follow-Ups

• A set of sorted tuples also dedupes — mention it, then explain why pointer-skipping avoids the extra hashing and allocation.
• Be ready for 4Sum: one more outer loop, same inner machinery. State the general k-sum recursion.
• Interviewers often ask why sorting is acceptable here (O(n log n) is under the O(n²) total) — have that ready.

That wraps part 7 of the Two Pointers arc. The full Technical Interview category maps all one hundred questions to the nine patterns that dominate FAANG screens — work through an arc end to end and the next unseen variant will feel familiar.