LeetCode 15. 3Sum

Description

https://leetcode.com/problems/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.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []

Constraints:

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

Explanation

Three sum is a follow-up question for two sum.

For three sum, we are going to find all possible triplets which meet the following criteria:

num1 + num2  + num3 = 0

Just make some modification on equation: num1 + num2 = -num3, it becomes a two sum problem: num1 + num2 = target, where target = -num3.

Therefore, for each value in the input array, we can use the two sum approach to search the remaining array for two elements whose sum = -value.

Java Solution

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> results = new ArrayList<>();
        
        if (nums == null || nums.length < 3) {
            return results;
        }
        
        Arrays.sort(nums);
        
        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            int target = -nums[i];
            int left = i + 1;
            int right = nums.length - 1;
            
            twoSumHelper(nums, results, target, left, right);
        }        
        
        return results;
    }
    
    private void twoSumHelper(int[] nums, List<List<Integer>> results, int target, int left, int right) {
        while (left < right) {
            if (nums[left] + nums[right] == target) {
                List<Integer> triplet = new ArrayList<>();                
                triplet.add(-target);
                triplet.add(nums[left]);
                triplet.add(nums[right]);
                results.add(triplet);
                
                left++;
                right--;
                
                while (left < right && nums[left] == nums[left - 1]) {
                    left++;
                }
                while (left < right && nums[right] == nums[right + 1]) {
                    right--;
                }
            } else if (nums[left] + nums[right] > target) {
                right--;
            } else if (nums[left] + nums[right] < target) {
                left++;                
            }
        }        
    }
}

Python Solution

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:

        def twosum_helper(target, results, nums, left, right, num):                
            while left < right:
                if nums[left] + nums[right] < target:
                    left += 1                    
                elif nums[left] + nums[right] > target:
                    right -= 1
                else:
                    triplet = [num, nums[left], nums[right]]
                    if triplet not in results:
                        results.append(triplet)
                    left += 1
                            
        results = []
        if len(nums) < 3:
            return results

        nums = sorted(nums)
                
        for i, num in enumerate(nums):
            target = -num
            twosum_helper(target, results, nums, i + 1, len(nums) - 1, num)
                        
        return results

• Time complexity: O(N^2).
• Space complexity: O(N).