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15. 3Sum

mediumAsked at Akamai

Find all unique triplets in an array that sum to zero. Akamai asks 3Sum to assess whether candidates can layer sorting on top of a two-pointer scan and handle duplicate pruning cleanly — the same deduplication discipline required in log-aggregation pipelines at scale.

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

Source citations

Public interview reports confirming this problem appears in Akamai loops.

  • Glassdoor (2026-Q1)— Akamai onsite reports list 3Sum as a common medium-difficulty coding question in algorithm rounds.
  • Blind (2025-10)— Multiple Akamai interview threads cite 3Sum as a canonical two-pointer problem asked in second-round interviews.

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. 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]]

Explanation: Two unique triplets sum to zero. Duplicates are pruned.

Example 2

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

Explanation: No triplet sums to zero.

Example 3

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

Explanation: The only triplet.

Approaches

1. Sort + two-pointer

Sort the array. For each element nums[i] (fixed as the first of the triplet), use a two-pointer scan of the remaining suffix to find pairs summing to -nums[i]. Skip duplicates at both the outer and inner levels.

Time
O(n²)
Space
O(1) excluding output
function threeSum(nums) {
  nums.sort((a, b) => a - b);
  const result = [];
  for (let i = 0; i < nums.length - 2; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) continue; // skip duplicate i
    let left = i + 1;
    let 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(n²) time, O(1) extra space. The sorting step is what enables two-pointer and deduplication. Early-exit when nums[i] > 0 is a worthwhile optimization to mention.

Akamai-specific tips

Akamai will probe the deduplication logic. Explain it before writing: 'After sorting, if nums[i] equals nums[i-1], the entire two-pointer sweep would produce identical triplets — so I skip. Same for the inner pointers after recording a result.' Also add the early-exit: 'If nums[i] > 0, all elements to the right are >= 0, so no triplet can sum to zero — we can break.'

Common mistakes

  • Skipping duplicates with nums[i] === nums[i+1] instead of nums[i-1] — skips valid triplets at the start of a duplicate run.
  • Forgetting to advance both pointers after recording a match — infinite loop on arrays with many equal elements.
  • Not sorting first and trying a hash-based dedup — far more complex and error-prone.

Follow-up questions

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

  • 3Sum Closest (LC 16) — return the triplet sum closest to a target value.
  • 4Sum (LC 18) — extend to four elements; generalize to k-sum recursively.
  • How would the complexity change if you needed all unique k-tuples summing to zero?

Solve it now

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Output

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FAQ

Why does sorting enable deduplication?

After sorting, identical values are adjacent. A simple comparison with the previous element is sufficient to detect and skip a duplicate pivot — no hash set needed.

Can we do better than O(n²)?

The best known algorithm for 3Sum is O(n²). A sub-quadratic solution would imply a breakthrough in computational geometry (3SUM-hardness). Mention this if the interviewer asks.

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