3Sum 问题 - Leet 码 - 超出时间限制

Question

I am trying to solve the 3Sum problem on Leetcode ( https://leetcode.com/problems/3sum/ ). the logic I am using is:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        k = list()
        l = len(nums)
        s= set()
        for idx,val in enumerate(nums):
            for j in range(idx+1,l):
                val2 = nums[j]
                temp = 0 - (val + val2)
                if temp in nums[j+1:]:
                    print(s)
                    if val not in k or val2 not in k or temp not in k:
                        k.append([val,val2,temp])
                        s.update([val,val2,temp])
        return k

In order to not add duplicate lists (lists with same elements) in the result, I am pushing all unique values to a set. Then checking if the set doesn't contain at least one of the three elements before adding to the list. I think this should prevent me from adding duplicate lists. Please correct me if I am wrong.

But the output result I see is :

Your input
    [-1,0,1,2,-1,-4]
stdout
    set()
    {0, 1, -1}
    {0, 1, 2, -1}
Output
    [[-1,0,1],[-1,2,-1],[0,1,-1]]
Expected
    [[-1,-1,2],[-1,0,1]]

I don't understand why [0,1,-1] is getting added to my result even though the set s contains all three elements 0,1,-1 . Where am I going wrong?

EDIT

Using the answers posted below, the final logic I built is:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        k = list()
        l = len(nums)
        for idx,val in enumerate(nums):
            for j in range(idx+1,l):
                val2 = nums[j]
                temp = 0 - (val + val2)
                if temp in nums[j+1:]:
                    item = sorted([val,val2,temp])
                    if item not in k:
                        k.append(item)
        return k

However, when I run this, time limit is exceeding for a large input. I get the error message Time Limit Exceeded . 311/313 testcases passed and 2 of them failed.

Is there a way to improvise the runtime for this logic?

Answer 1

Take a look at this line of your code

if val not in k or val2 not in k or temp not in k:

Here k is lists of lists and val,val2,temp are int . So this is definitely evaluate to True and you always append to the list.

Also you must first sort the lists before adding to the k , so that [-1,0,1] and [0,1,-1] are identified as same.

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        k = list()
        l = len(nums)
        s= set()
        for idx,val in enumerate(nums):
            for j in range(idx+1,l):
                val2 = nums[j]
                temp = 0 - (val + val2)
                if temp in nums[j+1:]:
                    print(s)
                    res = sorted([val,val2,temp])
                    if res not in k:
                        k.append(res)
                        s.update(res)
        return k

Result is:

set()
{0, 1, -1}
{0, 1, 2, -1}
[[-1, 0, 1], [-1, 2, -1]]

Answer 2

Sort your list, so that all solution lists will be in order. Then add only when the new list is not in the solution set -- your current test is useless, as it checks scalars against lists, which will always result in "I need this solution".

nums = [-1,0,1,2,-1,-4]
nums.sort()
print(nums)
k = list()
l = len(nums)
s= set()
old_val2 = None
for idx, val in enumerate(nums):
    for j in range(idx+1,l):
        val2 = nums[j] 
        temp = 0 - (val + val2)
        if temp in nums[j+1:]:
            print(s)
            # if val not in k or val2 not in k or temp not in k:
            if [val, val2, temp] not in k:
                k.append([val,val2,temp])
                s.update([val,val2,temp])
print(k)

Output:

[-4, -1, -1, 0, 1, 2]
set()
{2, -1}
{0, 1, 2, -1}
[[-1, -1, 2], [-1, 0, 1]]