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19. 3Sum

mediumAsked at Square

Find every trio of transaction adjustments in a ledger that net to zero — Square's reconciliation engine flags three-way payment splits that cancel out, which can signal money-laundering patterns in its fraud-detection pipeline.

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

Problem

Given an integer array nums, return all triplets [nums[i], nums[j], nums[k]] such that i, j, and k are distinct indices 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]]

Example 2

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

Example 3

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

Approaches

1. Brute force

Check all O(n^3) triplets; deduplicate with a set. Times out on large inputs.

Time
O(n^3)
Space
O(n)
function threeSum(nums) {
  const res = new Set();
  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);
          res.add(JSON.stringify(trip));
        }
      }
    }
  }
  return [...res].map(JSON.parse);
}

Tradeoff:

2. Sort + two pointers

Sort the array, then for each element use two pointers to find complementary pairs. Skip duplicates by advancing pointers past repeated values. O(n^2) time, no hash set needed.

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

Tradeoff:

Square-specific tips

Square's fraud team works with signed integer ledger deltas daily, so they expect you to reason about the duplicate-skip logic under pressure. Trace through [-1,-1,0,1,2] on the whiteboard and show exactly which pointer moves skip the second -1 — that real-data trace is what separates candidates who memorized the pattern from those who own it.

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Output

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