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16. 3Sum

mediumAsked at Ramp

Find all unique triplets in the array that sum to zero.

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

Problem

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i, j, k are distinct 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 = [0,0,0]
Output
[[0,0,0]]

Approaches

1. Brute force

Three nested loops plus a Set to dedupe sorted triplets.

Time
O(n^3)
Space
O(n)
function threeSum(nums) {
  const set = 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)
          set.add([nums[i], nums[j], nums[k]].sort((a,b)=>a-b).join(','));
  return [...set].map(s => s.split(',').map(Number));
}

Tradeoff:

2. Sort + two pointers per anchor

Sort the array; for each anchor i, two-pointer scan the suffix for pairs that sum to -nums[i]. Skip duplicates at every level.

Time
O(n^2)
Space
O(1)
function threeSum(nums) {
  nums.sort((a, b) => a - b);
  const out = [];
  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 s = nums[i] + nums[l] + nums[r];
      if (s === 0) {
        out.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 (s < 0) l++;
      else r--;
    }
  }
  return out;
}

Tradeoff:

Ramp-specific tips

Ramp's vendor-matching layer runs an analogous anchor-and-pair pattern across invoice line items, so two-pointer dedup discipline is a high-bonus signal here.

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