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11. 3Sum

mediumAsked at Autodesk

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 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 triple loop

Try every triple and store unique sorted triples.

Time
O(n^3)
Space
O(n)
for (let i=0;i<n;i++)
  for (let j=i+1;j<n;j++)
    for (let k=j+1;k<n;k++)
      if (a[i]+a[j]+a[k]===0) add(sorted([a[i],a[j],a[k]]));

Tradeoff:

2. Sort + two-pointer

Sort, fix one element, and use two pointers to find pairs summing to the negative target. Skip duplicates to keep results unique.

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:

Autodesk-specific tips

Autodesk likes seeing duplicate-skipping discipline because the same idea underpins canonicalizing duplicate triangles in mesh data structures.

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