3SumMedium

Given an array of integers, find all unique triplets [a,b,c] such that a + b + c = 0. The output must not contain duplicate triplets.

Examples

Input: nums = [-1,0,1,2,-1,-4] → Output: [[-1,-1,2],[-1,0,1]]

Approach

Three nested loops with a Set to dedupe triplets works, but it's O(n³) and the dedupe step is clunky. Sorting the array first unlocks something better: fix one number, then use the two-pointer technique on the remaining two — since the array is sorted, you can slide left/right pointers toward each other based on whether the current sum is too big or too small, exactly like Two Sum on a sorted array. Sorting also makes skipping duplicate values trivial, since equal values end up adjacent.

Complexity — best & worst case

Brute force — time O(n³)

Optimal — time O(n²), best and worst case are the same — one O(n) two-pointer scan per anchor, n anchors

Optimal — space O(log n) to O(n), depending on the sort's implementation

Code

// Given an array of integers, find all unique triplets that sum to
// zero. No duplicate triplets in the output.

// --- Brute force: three nested loops, dedupe with a Set of sorted keys ---
function threeSumBrute(nums) {
  const results = new Set();
  const n = nums.length;
  for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {
      for (let k = j + 1; k < n; k++) {
        if (nums[i] + nums[j] + nums[k] === 0) {
          const triplet = [nums[i], nums[j], nums[k]].sort((a, b) => a - b);
          results.add(JSON.stringify(triplet));
        }
      }
    }
  }
  return [...results].map((s) => JSON.parse(s));
}

// --- Optimal: sort once, fix one number, two-pointer the rest ---
// Sorting lets us skip duplicates cheaply and use the two-pointer
// pattern (see two-pointers.js) on the remaining two numbers.
function threeSum(nums) {
  const sorted = [...nums].sort((a, b) => a - b);
  const results = [];

  for (let i = 0; i < sorted.length - 2; i++) {
    if (i > 0 && sorted[i] === sorted[i - 1]) continue; // skip duplicate anchors

    let left = i + 1;
    let right = sorted.length - 1;

    while (left < right) {
      const sum = sorted[i] + sorted[left] + sorted[right];
      if (sum === 0) {
        results.push([sorted[i], sorted[left], sorted[right]]);
        left++;
        right--;
        while (left < right && sorted[left] === sorted[left - 1]) left++; // skip dupes
        while (left < right && sorted[right] === sorted[right + 1]) right--;
      } else if (sum < 0) {
        left++; // need a bigger sum
      } else {
        right--; // need a smaller sum
      }
    }
  }
  return results;
}

// --- Example Usage ---
console.log(threeSum([-1, 0, 1, 2, -1, -4]));
// [[-1,-1,2], [-1,0,1]]
console.log(threeSumBrute([-1, 0, 1, 2, -1, -4]));
// same triplets, unordered