3Sum

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

Contents

Approach 1: Brute Force O(n³)

Try all combinations of three indices and collect those that sum to zero. Use a Set to deduplicate.

Java
public List<List<Integer>> threeSum(int[] nums) {
    Set<List<Integer>> result = new HashSet<>();
    int n = nums.length;
    for (int i = 0; i < n - 2; i++) {
        for (int j = i + 1; j < n - 1; j++) {
            for (int k = j + 1; k < n; k++) {
                if (nums[i] + nums[j] + nums[k] == 0) {
                    List<Integer> triplet = Arrays.asList(nums[i], nums[j], nums[k]);
                    Collections.sort(triplet);
                    result.add(triplet);
                }
            }
        }
    }
    return new ArrayList<>(result);
}
// Time: O(n³), Space: O(n)

Approach 2: Sort + Two Pointers O(n²)

Fix one element at index i, then run a two-pointer search for the remaining two elements in the sorted array. Skip duplicate values at each level to avoid duplicate triplets.

Implementation steps:
Java
public List<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> result = new ArrayList<>();

    for (int i = 0; i < nums.length - 2; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicate pivot
        if (nums[i] > 0) break; // sorted: no triplet can sum to 0

        int left = i + 1, right = nums.length - 1;
        while (left < right) {
            int sum = nums[i] + nums[left] + nums[right];
            if (sum == 0) {
                result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                while (left < right && nums[left]  == nums[left  + 1]) left++;  // skip dups
                while (left < right && nums[right] == nums[right - 1]) right--; // skip dups
                left++;
                right--;
            } else if (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

Complexity Analysis
ApproachTimeSpace
Brute ForceO(n³)O(n)
Sort + Two PointersO(n²)O(1) extra (sort is in-place)