#15 - 3Sum

MediumArray / String

Given an integer array nums, return all unique 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.

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

Solution Space

Try it

/**
 * Returns all unique triplets that sum to zero.
 *
 * @param nums Array of integers (may contain duplicates)
 * @returns Array of unique triplets summing to 0
 */
export function threeSum(nums: number[]): number[][] {
  const result: number[][] = [];
  nums.sort((a, b) => a - b);

  for (let i = 0; i < nums.length - 2; i++) {
    // Skip duplicate values for the first element
    if (i > 0 && nums[i] === nums[i - 1]) continue;

    // Early termination: smallest possible sum already > 0
    if (nums[i] > 0) break;

    let left = i + 1;
    let 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]]);
        // Skip duplicates for both pointers
        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;
}

Approach: sort + two-pointer – sort the array, iterate through it fixing one element, and use two pointers to find pairs that sum to the negative of the fixed element.

Time: O(n²) – outer loop O(n) × inner two-pointer scan O(n).

Space: O(1) extra – sorting is in-place; output space is not counted.

Alternative approaches:

Approach	Idea	Time	Space
Hash-set	For each pair, check if -(a+b) exists in a set; use a Set<string> to deduplicate triplets.	O(n²)	O(n)
Brute force	Check all triples i < j < k.	O(n³)	O(1)

Why sort + two-pointer is preferred:

• Sorting makes duplicate skipping trivial with simple === checks on adjacent elements.
• O(1) extra space vs. the hash-set approach which needs bookkeeping to avoid duplicate triplets.

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