Problem 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.
Examples
Example 1: 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: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0. Example 3: Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.
Python Solution
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = []
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left, right = i + 1, len(nums) - 1
while left < right:
total = nums[i] + nums[left] + nums[right]
if total == 0:
result.append([nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
elif total < 0:
left += 1
else:
right -= 1
return resultJava Solution
class Solution {
public List> threeSum(int[] nums) {
Arrays.sort(nums);
List> result = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int total = nums[i] + nums[left] + nums[right];
if (total == 0) {
result.add(Arrays.asList(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 (total < 0) {
left++;
} else {
right--;
}
}
}
return result;
}
} C++ Solution
class Solution {
public:
vector> threeSum(vector& nums) {
sort(nums.begin(), nums.end());
vector> result;
for (int i = 0; i < nums.size() - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1;
int right = nums.size() - 1;
while (left < right) {
int total = nums[i] + nums[left] + nums[right];
if (total == 0) {
result.push_back({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 (total < 0) {
left++;
} else {
right--;
}
}
}
return result;
}
}; JavaScript Solution
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(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;
}
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
const total = nums[i] + nums[left] + nums[right];
if (total === 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 (total < 0) {
left++;
} else {
right--;
}
}
}
return result;
};C# Solution
public class Solution {
public IList> ThreeSum(int[] nums) {
Array.Sort(nums);
var result = new List>();
for (int i = 0; i < nums.Length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1;
int right = nums.Length - 1;
while (left < right) {
int total = nums[i] + nums[left] + nums[right];
if (total == 0) {
result.Add(new List { 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 (total < 0) {
left++;
} else {
right--;
}
}
}
return result;
}
} Complexity Analysis
- Time Complexity: O(n²) where n is the length of the input array
- Space Complexity: O(1) as we only use a constant amount of extra space
Solution Explanation
This solution uses the two-pointer technique:
- First, we sort the array to handle duplicates and make the two-pointer approach work
- For each number (i), we:
- Use two pointers (left and right) to find pairs that sum to -nums[i]
- Skip duplicates to avoid duplicate triplets
- Move pointers based on the current sum
Key points:
- Sorting helps us avoid duplicates and makes the two-pointer approach efficient
- We skip duplicate numbers to avoid duplicate triplets
- The two-pointer technique reduces the time complexity from O(n³) to O(n²)
- We handle edge cases (empty array, all zeros) properly