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.Solution in Python:
To solve the problem of finding all unique triplets in an array that sum up to zero, we can use a two-pointer approach after sorting the array. This approach ensures an efficient solution with O(n2) time complexity.
Python
from typing import List
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
# Sort the array
nums.sort()
result = []
# Iterate through the array
for i in range(len(nums) - 2):
# Skip duplicate elements to avoid duplicate triplets
if i > 0 and nums[i] == nums[i - 1]:
continue
# Initialize two pointers
left, right = i + 1, len(nums) - 1
while left < right:
total = nums[i] + nums[left] + nums[right]
if total == 0:
# Found a triplet
result.append([nums[i], nums[left], nums[right]])
# Skip duplicates for left pointer
while left < right and nums[left] == nums[left + 1]:
left += 1
# Skip duplicates for right pointer
while left < right and nums[right] == nums[right - 1]:
right -= 1
# Move both pointers
left += 1
right -= 1
elif total < 0:
# Move left pointer to the right to increase the sum
left += 1
else:
# Move right pointer to the left to decrease the sum
right -= 1
return result
# Example usage:
sol = Solution()
print(sol.threeSum([-1, 0, 1, 2, -1, -4])) # Output: [[-1, -1, 2], [-1, 0, 1]]
print(sol.threeSum([0, 1, 1])) # Output: []
print(sol.threeSum([0, 0, 0])) # Output: [[0, 0, 0]]Explanation:
- Sorting:
- The array is sorted to make it easier to avoid duplicates and to use the two-pointer technique efficiently.
- Main Loop:
- The loop iterates through the array, fixing one element
nums[i]at a time. - If
nums[i]is the same as the previous element, it is skipped to avoid duplicate triplets.
- The loop iterates through the array, fixing one element
- Two-Pointer Technique:
- Two pointers,
leftandright, are initialized.leftstarts right afteri, andrightstarts at the end of the array. - The sum of the triplet
nums[i] + nums[left] + nums[right]is calculated. - If the sum is zero, the triplet is added to the result list.
- If the sum is less than zero, the
leftpointer is moved to the right to increase the sum. - If the sum is greater than zero, the
rightpointer is moved to the left to decrease the sum.
- Two pointers,
- Avoiding Duplicates:
- After finding a valid triplet, the
leftandrightpointers are moved to skip over any duplicate values to ensure that each triplet is unique.
- After finding a valid triplet, the
This solution is efficient and ensures that all unique triplets are found with a time complexity of O(n2) making it suitable for the given constraints.
Solution in Javascript:
JavaScript
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(nums) {
// Sort the array
nums.sort((a, b) => a - b);
let result = [];
// Iterate through the array
for (let i = 0; i < nums.length - 2; i++) {
// Skip duplicate elements to avoid duplicate triplets
if (i > 0 && nums[i] === nums[i - 1]) continue;
// Initialize two pointers
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
let total = nums[i] + nums[left] + nums[right];
if (total === 0) {
// Found a triplet
result.push([nums[i], nums[left], nums[right]]);
// Skip duplicates for left pointer
while (left < right && nums[left] === nums[left + 1]) left++;
// Skip duplicates for right pointer
while (left < right && nums[right] === nums[right - 1]) right--;
// Move both pointers
left++;
right--;
} else if (total < 0) {
// Move left pointer to the right to increase the sum
left++;
} else {
// Move right pointer to the left to decrease the sum
right--;
}
}
}
return result;
};
// Example usage:
console.log(threeSum([-1, 0, 1, 2, -1, -4])); // Output: [[-1, -1, 2], [-1, 0, 1]]
console.log(threeSum([0, 1, 1])); // Output: []
console.log(threeSum([0, 0, 0])); // Output: [[0, 0, 0]]Solution in Java:
Java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// Sort the array
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
// Iterate through the array
for (int i = 0; i < nums.length - 2; i++) {
// Skip duplicate elements to avoid duplicate triplets
if (i > 0 && nums[i] == nums[i - 1]) continue;
// Initialize two pointers
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int total = nums[i] + nums[left] + nums[right];
if (total == 0) {
// Found a triplet
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
// Skip duplicates for left pointer
while (left < right && nums[left] == nums[left + 1]) left++;
// Skip duplicates for right pointer
while (left < right && nums[right] == nums[right - 1]) right--;
// Move both pointers
left++;
right--;
} else if (total < 0) {
// Move left pointer to the right to increase the sum
left++;
} else {
// Move right pointer to the left to decrease the sum
right--;
}
}
}
return result;
}
// Example usage
public static void main(String[] args) {
Solution sol = new Solution();
List<List<Integer>> result = sol.threeSum(new int[]{-1, 0, 1, 2, -1, -4});
for (List<Integer> triplet : result) {
System.out.println(triplet);
}
}
}Solution in C#:
C#
using System;
using System.Collections.Generic;
public class Solution {
public IList<IList<int>> ThreeSum(int[] nums) {
// Sort the array
Array.Sort(nums);
IList<IList<int>> result = new List<IList<int>>();
// Iterate through the array
for (int i = 0; i < nums.Length - 2; i++) {
// Skip duplicate elements to avoid duplicate triplets
if (i > 0 && nums[i] == nums[i - 1]) continue;
// Initialize two pointers
int left = i + 1;
int right = nums.Length - 1;
while (left < right) {
int total = nums[i] + nums[left] + nums[right];
if (total == 0) {
// Found a triplet
result.Add(new List<int> { nums[i], nums[left], nums[right] });
// Skip duplicates for left pointer
while (left < right && nums[left] == nums[left + 1]) left++;
// Skip duplicates for right pointer
while (left < right && nums[right] == nums[right - 1]) right--;
// Move both pointers
left++;
right--;
} else if (total < 0) {
// Move left pointer to the right to increase the sum
left++;
} else {
// Move right pointer to the left to decrease the sum
right--;
}
}
}
return result;
}
}