Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

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: []

Example 3:

• Input: nums = [0,0,0]
• Output: [[0,0,0]]

Constraints:

• 3 ≤ nums.length ≤ 3000
• -10⁵ ≤ nums[i] ≤ 10⁵

Why this problem exists: 3Sum is a classic problem that extends the two-sum concept and tests your ability to avoid O(n³) solutions through sorting and two-pointer technique.

What it teaches: • Sorting + two-pointer combination • Avoiding duplicate triplets • Reducing complexity from O(n³) to O(n²)

Interview relevance: One of the most frequently asked medium-difficulty problems. Tests combinatorial thinking and optimization.

Pattern: Fixed + Two Pointers

Fix one element with a loop, then use two pointers (left, right) on the remaining subarray to find pairs that sum to -(fixed value). Sort first to enable two-pointer search and duplicate handling.

When to use this pattern: • Finding k-sum combinations • Any problem where sorting + two pointers can reduce complexity • Finding pairs that satisfy a condition

graph LR
    L["Left Pointer (L)"] --> Array["Input Array / String"]
    R["Right Pointer (R)"] --> Array
    Array --> Condition{"Check Window Condition"}
    Condition -- "Expand R" --> R
    Condition -- "Shrink L" --> L
    Condition -- "Valid State" --> Max["Update Max / Subarray Result"]

function threeSum(nums) {
  const result = [];
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      for (let k = j + 1; k < nums.length; k++) {
        if (nums[i] + nums[j] + nums[k] === 0) {
          result.push([nums[i], nums[j], nums[k]]);
        }
      }
    }
  }
  return result;
}

• Time Complexity: O(n³)
• Space Complexity: O(n)
• Explanation: Check every possible triplet — O(n³) time. Also includes duplicates.

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; // skip duplicates

    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;
}

• Time Complexity: O(n²)
• Space Complexity: O(1) excluding output
• Explanation: Sort the array (O(n log n)). Fix one element, then use two pointers on the remaining subarray. Skip duplicates to avoid repeat triplets. O(n²) total.

1. Initialize State: Setup necessary pointers, dynamic programming arrays, or hash maps.
2. Iterate & Evaluate: Process the input according to the boundary conditions.
3. Update & Return: Compute the optimal answer and return early or at termination.

How to explain:

1. Sort is essential for both two-pointer and duplicate handling
2. Fix one element (skip duplicates). For each, use two pointers on the rest
3. Move pointers based on sum: too low → left++, too high → right—
4. Skip duplicates when found

Follow-ups: • “What about 3Sum closest?” → Track the closest sum instead of exact match • “What about 4Sum?” → Add another nested loop (O(n³)) or use the same pattern recursively • “What if the array is already sorted?” → Skip the sorting step, O(n²) directly

1. Sort the array first — enables two-pointer approach and duplicate handling.
2. Fix one element, then use two pointers on the remaining elements to find pairs summing to -(fixed value).
3. Skip duplicate values to avoid duplicate triplets in the result.

👉 Solve this problem interactively in the DSA Lab