14. 3Sum
mediumAsked at ByteDanceReturn all unique triplets summing to zero — ByteDance uses it to test sort + two-pointer scaffolding and de-duplication discipline.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an integer array nums, return every unique triplet [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 = [0,0,0][[0,0,0]]Approaches
1. Triple loop with set dedup
Try every triple, sort each tuple, and dedup with a string set.
- Time
- O(n^3)
- Space
- O(n^2)
// nested loops -> push sorted triples into a Set keyed by JSONTradeoff:
2. Sort then two-pointer scan
Sort the array, fix one anchor, then two-pointer scan the rest. Skip equal neighbors to avoid duplicate triplets.
- 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:
ByteDance-specific tips
ByteDance interviewers grade hard on the dedup skip lines — sloppy skips become bugs in their content-fingerprint pipeline, so they want them explicitly named.
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