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Medium
3Sum
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
Input:nums = [-1,0,1,2,-1,-4]
Output:[[-1,-1,2],[-1,0,1]]
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].
Input:nums = [0,1,1]
Output:[]
The only possible triplet does not sum up to 0.
Constraints
3 <= nums.length <= 3000-10^5 <= nums[i] <= 10^5
Approach
Use three nested loops to check every possible triplet combination. To ensure no duplicate triplets are added, sort the triplets and add them to a hash set before returning the result.
Complexity Analysis
Time Complexity
O(n^3)
Space Complexity
O(n)
This approach is highly inefficient and will result in Time Limit Exceeded (TLE) for large arrays.
Solution.java
class Solution { public List<List<Integer>> threeSum(int[] nums) { Set<List<Integer>> res = new HashSet<>(); for (int i = 0; i < nums.length - 2; i++) { for (int j = i + 1; j < nums.length - 1; j++) { for (int k = j + 1; k < nums.length; k++) { if (nums[i] + nums[j] + nums[k] == 0) { List<Integer> triplet = Arrays.asList(nums[i], nums[j], nums[k]); Collections.sort(triplet); res.add(triplet); } } } } return new ArrayList<>(res); }}