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35. 3Sum

mediumAsked at Ola

Find 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, k are distinct and nums[i] + nums[j] + nums[k] == 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 triples

Three nested loops; dedup with a stringified set.

Time
O(n^3)
Space
O(n)
const set = 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) set.add([nums[i],nums[j],nums[k]].sort().join(','));
return [...set].map(s=>s.split(',').map(Number));

Tradeoff:

2. Sort then two-pointer

Sort once; for each pivot run two pointers inward; 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:

Ola-specific tips

Ola probes how you handle duplicates carefully; tie it to grouping three drivers whose net surplus zeros out across regions.

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