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16. 3Sum

mediumAsked at Squarespace

Find every unique triplet that sums to zero; Squarespace uses it to test whether you can combine sorting with the two-pointer pattern and de-duplicate cleanly.

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

Problem

Given an integer array nums, return all unique triplets [a, b, c] such that a + b + c = 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
[]

Approaches

1. Brute force triples

Enumerate every triple and dedupe results via a stringified set.

Time
O(n^3)
Space
O(n^2)
const seen=new Set(),res=[];
for(let i=0;i<n;i++) for(let j=i+1;j<n;j++) for(let k=j+1;k<n;k++)
  if(nums[i]+nums[j]+nums[k]===0){
    const t=[nums[i],nums[j],nums[k]].sort((a,b)=>a-b);
    const k2=t.join(','); if(!seen.has(k2)){seen.add(k2);res.push(t);}
  }
return res;

Tradeoff:

2. Sort then two-pointer scan

Sort the array, fix one number, and use two pointers to find the matching pair while skipping duplicates.

Time
O(n^2)
Space
O(1)
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 l = i + 1, r = nums.length - 1;
    while (l < r) {
      const s = nums[i] + nums[l] + nums[r];
      if (s === 0) {
        res.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 res;
}

Tradeoff:

Squarespace-specific tips

Squarespace evaluators expect the dedupe-skip on the outer and inner pointers and they like a note connecting it to dedupe logic in the media library upload pipeline.

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Output

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