3sum

Solution 1

• Time complexity: O(n³)
• Space complexity: O(n)

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        
        n = len(nums)
        if n < 3: return []
        res = []
        
        sorted_nums = list(sorted(nums))
        
        for i in range(n-2):
            for j in range(i+1, n-1):
                for k in range(j+1, n):
                    if sorted_nums[i] + sorted_nums[j] + sorted_nums[k] == 0:
                        if [sorted_nums[i], sorted_nums[j], sorted_nums[k]] not in res:
                            res.append([sorted_nums[i], sorted_nums[j], sorted_nums[k]])

        return res
    

Solution 2

• Binary search (by sorting first)
  • Time complexity: O(n²)
  • Space complexity: O(n)

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        
        # Time complexity: O(n^2)
        # Space complexity: O(n)
        
        n = len(nums)
        if n < 3: return []
        res = []
        
        sorted_nums = list(sorted(nums))
        
        for i in range(n-2):
            
            if i > 0 and sorted_nums[i-1] == sorted_nums[i]: # Skip already found threeSum for a fixed sorted_nums[i]
                continue
                
            l, r = i+1, n-1
            
            while l < r:
                
                threeSum = sorted_nums[i] + sorted_nums[l] + sorted_nums[r] 
                
                if threeSum > 0:
                    r -= 1
                elif threeSum < 0:
                    l += 1
                else:
                    res.append([sorted_nums[i], sorted_nums[l], sorted_nums[r]])
                    l += 1
                    while sorted_nums[l] == sorted_nums[l-1] and l < r: # Find all unique threeSum for a fixed sorted_nums[i]  
                        l += 1
        return res