15. 3Sum
mediumAsked at KlarnaFind 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 the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, j != k, and the three values sum to zero. The result 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. Triple loop
Enumerate every triple and use a set of sorted keys to dedupe.
- Time
- O(n^3)
- Space
- O(n)
function threeSum(nums) {
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 + two-pointer
Sort the array, then for each i, run a two-pointer sweep across the remainder looking for the pair that completes a zero sum. Skip duplicates as you go.
- 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:
Klarna-specific tips
Klarna interviewers care more about how cleanly you skip duplicate triplets than about the asymptotic — sloppy dedupe is what they see in production risk-feature jobs that explode under cardinality.
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