15._3Sum_.py

"""Problem No.: 15
Problem:  3Sum
DS:  Hash map
Approach:
Date:  26/07/2023
Detailed Problem: 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.
"""
class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]: 
        result = []
        nums.sort()
        for i in range(len(nums)):
            if nums[i]>0:
                break
            if i ==0 or nums[i-1] != nums[i]:
                self.twoSum(nums,i,result)
        return result

    
    def twoSum(self,nums, i,result):
        low = i+1
        high = len(nums)-1
        while (low <high):
            sum = nums[i]+nums[low]+nums[high]
            if sum>0:
                high -=1               
            elif sum<0:
                low +=1
            else: 
                result.append([nums[i],nums[low],nums[high]])
                low +=1
                high -=1
                while (low<high )and nums[low]== nums[low-1]:
                    low +=1
                
               
"""
Time Complexity:O(n*2). twoSum is O(n), and we call it n times.
Sorting the array takes O(nlog⁡n), so overall complexity is O(nlog⁡n+n2). This is asymptotically equivalent to O(n2)
Space Complexity: O(n) for the hashset.
"""