17. 3Sum
mediumAsked at MercuryFind all unique triplets in an array that sum to zero.
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
nums = [-1,0,1,2,-1,-4][[-1,-1,2],[-1,0,1]]Example 2
nums = [0,1,1][]Approaches
1. Brute triple loop
Check every (i,j,k) triple and dedupe with a Set of sorted tuples.
- Time
- O(n^3)
- Space
- O(n)
const out=new Set();
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) 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 + two-pointer per anchor
Sort the array, fix each anchor, then close the remaining target with two pointers; skip duplicate anchors and duplicate inner moves.
- 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 sum = nums[i] + nums[l] + nums[r];
if (sum < 0) l++;
else if (sum > 0) r--;
else {
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--;
}
}
}
return res;
}Tradeoff:
Mercury-specific tips
Mercury reuses two-pointer narrative for reconciling triples of ledger entries — a clearing leg plus its inbound and outbound mirrors should sum to zero in a net-zero internal book.
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