3 Sum
AckoHHashedIn
SprinklrThe problem
Given an integer array nums. Return all triplets such that:
-
i != j, i != k, and j != k
-
nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets. One element can be a part of multiple triplets. The output and the triplets can be returned in any order.
Input: nums = [2, -2, 0, 3, -3, 5] Output: [[-2, 0, 2], [-3, -2, 5], [-3, 0, 3]] Explanation: nums[1] + nums[2] + nums[0] = 0 nums[4] + nums[1] + nums[5] = 0 nums[4] + nums[2] + nums[3] = 0
Input: nums = [2, -1, -1, 3, -1] Output: [[-1, -1, 2]] Explanation: nums[1] + nums[2] + nums[0] = 0 Note that we have used two -1s as they are separate elements with different indexes But we have not used the -1 at index 4 as that would create a duplicate triplet
Input: nums = [8, -6, 5, 4] (Give answer with the output and triplets sorted in ascending order)
- 1 <= nums.length <= 3000
- -104 <= nums[i] <= 104
cpp
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
}
};java
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
}
}python
class Solution:
def threeSum(self, nums: list) -> list[list]:javascript
class Solution {
threeSum(nums) {
}
}csharp
public class Solution {
public IList<IList<int>> ThreeSum(int[] nums) {
}
}go
func threeSum(nums []int) [][]int {
}Stuck? Show a way to structure it+
- 01Sort the values and fix each possible first element once.
- 02Search the remaining suffix with left and right pointers.
- 03After finding a triplet, skip duplicate values at all three positions.
- 04Use sorted-order bounds to stop early when no later triplet can sum to zero.
Reference answer
Then expect these follow-ups
How would you generalize this to k-sum?
Tests: recursion pattern
Why can the scan stop when the fixed value is positive?
Tests: sorted reasoning
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