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13. 3Sum

mediumAsked at Freshworks

Find every distinct triplet that sums to zero — Freshworks dresses it as balancing three SLA-credit ledgers to net 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

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. Brute force

Three nested loops, push every zero-sum triplet, then dedup via a Set of sorted strings.

Time
O(n^3)
Space
O(n)
const seen = new Set(), out = [];
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) {
    const key=[nums[i],nums[j],nums[k]].sort((a,b)=>a-b).join(',');
    if (!seen.has(key)) { seen.add(key); out.push(key.split(',').map(Number)); }
  }
return out;

Tradeoff:

2. Sort + two pointers

Sort. For each anchor i, run two pointers l = i+1, r = n-1 to find pairs summing to -nums[i]. Skip duplicates on i, l, and r.

Time
O(n^2)
Space
O(1) extra
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:

Freshworks-specific tips

Freshworks always asks about deduplication — narrate the three skip-duplicates branches (anchor, left, right) explicitly because that's where most candidates ship a wrong solution.

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Output

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