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.
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Input: nums = [0,1,1]
Output: []
Input: nums = [0,0,0]
Output: [[0,0,0]]
Constraints:
- 3 <= nums.length <= 3000
- -105 <= nums[i] <= 105
Contents
- Approach 1 — Brute Force O(n³)
- Approach 2 — Sort + Two Pointers O(n²)
- Complexity Analysis
Brute Force O(n³)
Try all combinations of three indices and collect those that sum to zero. Use a Set to deduplicate.
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)
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:
- Sort the array.
- Iterate i from 0 to n-3. Skip if nums[i] == nums[i-1].
- Set left = i+1, right = n-1.
- While left < right: if sum == 0, record; skip duplicates at left and right; adjust pointers.
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;
}
| Approach | Time | Space |
|---|
| Brute Force | O(n³) | O(n) |
| Sort + Two Pointers | O(n²) | O(1) extra (sort is in-place) |