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17. 3Sum

mediumAsked at Wise

Find every triplet in an array that sums to zero — Wise uses it as a three-leg peer pooling probe (two senders cover one receiver).

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 returned triplets must not contain duplicates.

Constraints

  • 3 <= nums.length <= 3000
  • -10^5 <= nums[i] <= 10^5
  • Return unique triplets only

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. Three nested loops with Set dedup

Brute every triple, store sorted triples in a Set.

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

Tradeoff:

2. Sort then two-pointer

Sort the array; fix nums[i] and use two pointers to find pairs that sum to -nums[i]. Skip duplicates at each level to keep results unique.

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:

Wise-specific tips

Wise grades the duplicate-skip discipline closely — peer matching in their FX engine must never propose the same three-leg pool twice, so a missed skip is a silent production bug.

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Output

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