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35. 3Sum

mediumAsked at Salesforce

Find all unique triplets in an array that sum to zero. Salesforce uses this to test sort + two-pointer + dedup composition — a real Salesforce engineer must combine all three.

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

Source citations

Public interview reports confirming this problem appears in Salesforce loops.

  • Glassdoor (2026-Q1)— Recurring on Salesforce backend onsites — they grade dedup carefully.
  • Blind (2025-09)— Specifically tests the sort + two-pointer hybrid.

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 = []
Output
[]

Example 3

Input
nums = [0]
Output
[]

Approaches

1. Brute force triple loop

Try every triple; dedup with stringified-sorted-triple set.

Time
O(n^3)
Space
O(unique triplets)
function threeSum(nums) {
  const seen = new Set();
  const result = [];
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      for (let k = j + 1; k < nums.length; k++) {
        if (nums[i] + nums[j] + nums[k] === 0) {
          const trip = [nums[i], nums[j], nums[k]].sort((a, b) => a - b).join(',');
          if (!seen.has(trip)) { seen.add(trip); result.push(trip.split(',').map(Number)); }
        }
      }
    }
  }
  return result;
}

Tradeoff: Cubic. TLE on 3000 elements. Salesforce wants O(n^2).

2. Sort + fix one + two-pointer for the rest

Sort. For each i, two-pointer l/r in (i, n) to find pairs summing to -nums[i]. Skip duplicates at each level.

Time
O(n^2)
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;
    let l = i + 1, r = nums.length - 1;
    while (l < r) {
      const sum = nums[i] + nums[l] + nums[r];
      if (sum === 0) {
        result.push([nums[i], nums[l], nums[r]]);
        while (l < r && nums[l] === nums[l + 1]) l++;
        while (l < r && nums[r] === nums[r - 1]) r--;
        l++; r--;
      } else if (sum < 0) l++;
      else r--;
    }
  }
  return result;
}

Tradeoff: O(n^2) time. The three dedup checks (one for i, two for l and r) are essential and the #1 source of bugs.

Salesforce-specific tips

Salesforce grades this on dedup correctness — most candidates get the sum logic right but produce duplicate triplets. Bonus signal: explain why you must skip nums[i] duplicates and also skip nums[l]/nums[r] duplicates AFTER finding a triplet (otherwise [-2, 0, 0, 2, 2] over-emits).

Common mistakes

  • Skipping i duplicate at i+1 vs i-1 — using i+1 misses the FIRST duplicate.
  • Not skipping l and r duplicates after finding a triplet — emits duplicate solutions.
  • Using the brute-force dedup-via-string approach in O(n^3) instead of sort+two-pointer.

Follow-up questions

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

  • 3Sum Closest (LC 16).
  • 4Sum (LC 18).
  • 3Sum Smaller (LC 259) — count triplets summing to less than target.

Solve it now

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Output

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FAQ

Why skip i duplicates with `if (i > 0 && nums[i] === nums[i-1])`?

Because if nums[i] equals the previous, the two-pointer scan for (i, n) is a subset of the previous scan — same triplets, just shifted. Skipping avoids duplicates.

Why two more dedup loops after finding a triplet?

To skip over runs of equal values at l and r. Otherwise the next iteration finds the same triplet again.

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