Problem Overview & Intuition

Sorting the array in O(N log N) allows solving the remainder with the two-pointer technique in O(N²). For each element nums[i], we search for pairs in the remaining subarray that sum to -nums[i]. Duplicate skipping prevents redundant triplets.

Step-by-Step Approach

1. Sort nums in ascending order.
2. Loop i from 0 to nums.length - 3.
3. If i > 0 && nums[i] === nums[i - 1], skip to prevent duplicates.
4. Use left = i + 1 and right = nums.length - 1.
5. While left < right:
6. If sum === 0, add [nums[i], nums[left], nums[right]], advance both pointers while skipping duplicates.
7. If sum < 0, left++; if sum > 0, right--.
8. Return result.

Optimal Implementation (JavaScript)

function threeSum(nums) {
  nums.sort((a, b) => a - b);
  const result = [];
  for (let i = 0; i < nums.length - 2; 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) {
        result.push([nums[i], nums[left], nums[right]]);
        while (left < right && nums[left] === nums[left + 1]) left++;
        while (left < right && nums[right] === nums[right - 1]) right--;
        left++;
        right--;
      } else if (sum < 0) left++;
      else right--;
    }
  }
  return result;
}

Complexity Analysis