Three Sum

Problem Source: Leetcode

Problem 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.

tests

def test(fn):
    nums = [-1,0,1,2,-1,-4]
    expected =  [[-1,-1,2],[-1,0,1]]
    assert fn(nums) == expected 

    nums = [0,1,1]
    expected = []
    assert fn(nums) == expected 

    nums = [0,0,0]
    expected = [[0,0,0]]
    assert fn(nums) == expected 

Solution

This solution builds on Two Sum II - Input Array is Sorted . If this solution is confusing, revisit it after understanding that proble,

• Time Complexity: O(1)
• Space Complexity: O(nlogn)

from typing import List
def threeSum(nums: List[int]) -> List[List[int]]:
    nums.sort()
    res = []

    for i in range(len(nums)):
        if i>0 and nums[i] == nums[i-1]: continue

        l,r = i+1, len(nums)-1
        while l < r:
            curr_sum = nums[i] + nums[l] + nums[r]

            #Two Sum on sorted list
            if curr_sum < 0: # If too small
                l += 1 # make it bigger
            elif curr_sum > 0: # If too big
                r -= 1 # make it smaller
            else:
                res.append([nums[i],nums[l],nums[r]])
                l += 1 # move to next
                while nums[l] == nums[l - 1] and l < r:
                    l += 1 # keep incrementing if dupes
    return res

test(threeSum)