15. 3Sum

Problem Link

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:

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.

 

Constraints:

• 3 <= nums.length <= 3000
• -105 <= nums[i] <= 105

Solution

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Python3

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        N = len(nums)
        res = []
        nums.sort()
 
        for i, x in enumerate(nums):
            if i > 0 and nums[i] == nums[i - 1]: continue
 
            left, right = i + 1, N - 1
 
            while left < right:
                if nums[left] + nums[right] + x == 0:
                    res.append([x, 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 nums[left] + nums[right] + x > 0:
                    right -= 1
                else:
                    left += 1
        
        return res

C++

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        int n = nums.size();
        
        sort(nums.begin(), nums.end());
        
        for (int i = 0; i < n-2; i++){
            
            if (i > 0 && nums[i] == nums[i-1]){
                continue;
            }
            
            int l = i+1, r = n-1;
            
            while (l < r){
                int s = nums[i] + nums[l] + nums[r];
                
                if (s < 0)
                    l++;
                
                else if (s > 0)
                    r--;
                
                else{
                    vector<int> temp(3, 0);
                    temp[0] = nums[i];
                    temp[1] = nums[l];
                    temp[2] = nums[r];
                    res.push_back(temp);
                    
                    while (l < r && nums[l] == nums[l+1]) l++;
                    while (l < r && nums[r] == nums[r-1]) r--;
                    
                    l++;
                    r--;
                }
            }         
        }
        return res;
    }
};