17. 3Sum
mediumAsked at WiseFind 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] <= 10^5Return unique triplets only
Examples
Example 1
nums=[-1,0,1,2,-1,-4][[-1,-1,2],[-1,0,1]]Example 2
nums=[0,1,1][]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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