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15. 3Sum

mediumAsked at Linear

Find all unique triplets in an array that sum to zero. Linear asks this to see if you know the 'anchor + two-pointer' pattern and can handle duplicate elimination cleanly — a level above Two Sum that exposes real interview preparation.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Linear loops.

  • Glassdoor (2026-Q1)— Consistently cited in Linear SWE onsite reports as a core medium problem.
  • Blind (2025-12)— Multiple Linear interview threads list 3Sum as a recurring onsite question.

Problem

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.

Constraints

  • 3 <= nums.length <= 3000
  • -10^5 <= nums[i] <= 10^5

Examples

Example 1

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

Example 2

Input
nums = [0,1,1]
Output
[]

Example 3

Input
nums = [0,0,0]
Output
[[0,0,0]]

Approaches

1. Brute force three loops

Check every triplet combination. Use a set to deduplicate results.

Time
O(n^3)
Space
O(n)
function threeSum(nums) {
  const result = new Set();
  for (let i = 0; i < nums.length - 2; i++) {
    for (let j = i + 1; j < nums.length - 1; j++) {
      for (let k = j + 1; k < nums.length; k++) {
        if (nums[i] + nums[j] + nums[k] === 0) {
          const t = [nums[i], nums[j], nums[k]].sort((a, b) => a - b);
          result.add(JSON.stringify(t));
        }
      }
    }
  }
  return [...result].map(JSON.parse);
}

Tradeoff: O(n^3) — correct but unacceptably slow. Establish this as a baseline, then pivot immediately.

2. Sort + anchor + two-pointer (optimal)

Sort the array. For each anchor element nums[i], use two pointers to find pairs in the remaining window that sum to -nums[i].

Time
O(n^2)
Space
O(1)
function threeSum(nums) {
  nums.sort((a, b) => a - b);
  const result = [];
  for (let i = 0; i < nums.length - 2; i++) {
    if (nums[i] > 0) break;
    if (i > 0 && nums[i] === nums[i - 1]) continue;
    let left = i + 1, right = nums.length - 1;
    while (left < right) {
      const sum = nums[i] + nums[left] + nums[right];
      if (sum === 0) {
        result.push([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;
}

Tradeoff: O(n^2) time, O(1) extra space (excluding output). Sorting enables two-pointer and also makes deduplication easy — skip over duplicate anchor values and duplicate pair values after recording a result.

Linear-specific tips

Linear interviewers expect you to name the pattern upfront: 'Sort, then anchor + two-pointer.' Walk through duplicate-skipping logic verbally before coding it — candidates who code first and explain second often miss the edge cases. Also mention that sorting converts this into a solvable Two Sum variant.

Common mistakes

  • Not skipping duplicate anchor values — leads to duplicate triplets in the output.
  • Not skipping duplicate pointer values after recording a match — same issue.
  • Skipping duplicates in the wrong direction (e.g., checking nums[left] === nums[left + 1] before incrementing left).
  • Forgetting to break early when nums[i] > 0 — if the anchor is positive, no triplet can sum to zero.

Follow-up questions

An interviewer at Linear may pivot to one of these next:

  • 3Sum Closest (LC 16) — find the triplet closest to target, not exactly zero.
  • 4Sum (LC 18) — generalize to four elements (anchor + 3Sum).
  • What if you need all unique quadruplets summing to a target?

Solve it now

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Output

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FAQ

Why sort first?

Sorting lets you use two pointers (O(n) per anchor) instead of a hash set (O(n) per anchor but with extra space). It also makes deduplication trivial — just skip consecutive equal values.

Why break when nums[i] > 0?

After sorting, if the smallest anchor is positive, all three numbers are positive and can't sum to zero. No point continuing.

Can I use a hash set instead of two pointers?

Yes — fix i and j, then look up -(nums[i]+nums[j]) in a set. But deduplication is harder to get right. Two-pointer is cleaner and preferred for this problem.

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