3Sum
Description
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:
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Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
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Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
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Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
Solutions
Let’s walk through the problem-solving process in Java, starting from a basic brute-force solution and gradually improving performance and readability.
✅ Brute Force (Time Complexity: O(n³), Space: O(1))
This approach checks every possible triplet and filters valid ones. It’s not efficient but helps understand the problem structure.
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class Solution {
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);
}
}
This solution works but is too slow for large input sizes. Let’s do better.
🚀 Optimized Using Sorting + Two Pointers (Time: O(n²), Space: O(n) for result)
We sort the array and fix one number, then use two pointers to find pairs that complete the triplet. This is the most commonly accepted optimal approach.
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class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicates
int j = i + 1;
int k = nums.length - 1;
while (j < k) {
int total = nums[i] + nums[j] + nums[k];
if (total < 0) {
j++;
} else if (total > 0) {
k--;
} else {
res.add(Arrays.asList(nums[i], nums[j], nums[k]));
j++;
k--;
while (j < k && nums[j] == nums[j - 1]) j++; // skip duplicates
while (j < k && nums[k] == nums[k + 1]) k--; // skip duplicates
}
}
}
return res;
}
}
This solution is efficient and avoids duplicates cleanly using two-pointer logic after sorting.
🧩 Alternate with HashSet to Track Triplets (Time: O(n²), Space: O(n²))
We use a Set to track unique triplets and avoid manual duplicate removal.
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class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Set<List<Integer>> result = new HashSet<>();
Arrays.sort(nums);
int n = nums.length;
for (int i = 0; i < n - 2; i++) {
int j = i + 1;
int k = n - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[j], nums[k]));
j++;
k--;
} else if (sum < 0) {
j++;
} else {
k--;
}
}
}
return new ArrayList<>(result);
}
}
While this version is straightforward, using Set adds a bit of overhead, and you still sort beforehand. It’s a good balance between simplicity and performance.
Summary
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force (Backtracking) | O(n³) | O(1) | Simple but slow |
| Two Pointers | O(n) | O(m * n) | Best mix of speed and clarity |
| HashSet Version | O(n²) | O(n²) | Avoids duplicate logic manually |