Demo preview: These solutions pass the automated test suite but have not been human-reviewed. Treat trade-offs and prose as draft.

Mediumtwo-pointers

3Sum

Problem (restated)

Given an integer array nums, return all unique triplets [a,b,c] such that a + b + c = 0. Triplets must be unique as sets of values (order in the triplet does not create duplicates).

Intuition

Sort, then fix one index i and run two-sum with two pointers on the right side. Skip duplicates at each pointer to keep triplets unique.

Approaches

Sort + two pointers

Tested only

Time O(n²)Space O(1) extra (ignoring output)

Idea. Sort nums. For each i, left=i+1, right=n-1. Move pointers by sum vs 0. Skip duplicate values for i, left, and right.

Walkthrough. [-1,0,1,2,-1,-4] → sorted [-4,-1,-1,0,1,2]. Fix -1, find (-1,0,1) and (-1,-1,2).

Trade-offs. O(n²) is expected; hashing pairs works but duplicate handling is messier.

Solution

export function threeSum(nums: number[]): number[][] {
  nums.sort((a, b) => a - b);
  const res: number[][] = [];
  for (let i = 0; i < nums.length; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) continue;
    let left = i + 1, right = nums.length - 1;
    while (left < right) {
      const sum = nums[i]! + nums[left]! + nums[right]!;
      if (sum === 0) {
        res.push([nums[i]!, nums[left]!, nums[right]!]);
        left++; right--;
        while (left < right && nums[left] === nums[left - 1]) left++;
        while (left < right && nums[right] === nums[right + 1]) right--;
      } else if (sum < 0) left++;
      else right--;
    }
  }
  return res;
}

export function threeSum(nums: number[]): number[][] {
  nums.sort((a, b) => a - b);
  const res: number[][] = [];
  for (let i = 0; i < nums.length; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) continue;
    let left = i + 1, right = nums.length - 1;
    while (left < right) {
      const sum = nums[i]! + nums[left]! + nums[right]!;
      if (sum === 0) {
        res.push([nums[i]!, nums[left]!, nums[right]!]);
        left++; right--;
        while (left < right && nums[left] === nums[left - 1]) left++;
        while (left < right && nums[right] === nums[right + 1]) right--;
      } else if (sum < 0) left++;
      else right--;
    }
  }
  return res;
}

Template connection

Two pointers on a sorted array after fixing one element. classic 3Sum extension of the pair-sum template.

Reflection

Which pattern gave this away within 90 seconds?
What changed from the standard template?
What would break the current solution?