3 Python 中的求和问题 3、Leetcode 超过时间限制

Question

I tried solving the 3Sum problem on Leetcode in python 3 but it shows that the time limit has exceeded for my solution. 3Sum problem is as follows: Given array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:

[-1, 0, 1], [-1, -1, 2]. I have been trying to figure it for a long time but in vain. Also, I am new to this so please bear with me.

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]
        res =[]
        size = len(nums)
        nums.sort()
        for i in range(size-2):
         
            if i> 0 and nums[i] == nums[i-1]:
                continue

            low = i+1
            high = size - 1

            while(low<high):
                val = nums[low]+nums[high] + nums[i]
                if(val < 0)  or ((low < high) and nums[low] == nums[low+1]):

                    low = low + 1
                elif( val > 0) or ((low < high) and nums[high] == nums[high-1]):

                    high = high - 1
                else:
                    res.append( (nums[i], nums[low], nums[high]))
        return(res)

Can someone please tell me what's the mistake.

Answer 1

You have to do that O(N ^ 2). O(N ^ 3) is not acceptable:

from typing import List

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        unique_triplets = []
        nums.sort()
        for i in range(len(nums) - 2):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            lo = i + 1
            hi = len(nums) - 1
            while lo < hi:
                target_sum = nums[i] + nums[lo] + nums[hi]
                if target_sum < 0:
                    lo += 1
                if target_sum > 0:
                    hi -= 1
                if target_sum == 0:
                    unique_triplets.append((nums[i], nums[lo], nums[hi]))
                    while lo < hi and nums[lo] == nums[lo + 1]:
                        lo += 1
                    while lo < hi and nums[hi] == nums[hi - 1]:
                        hi -= 1
                    lo += 1
                    hi -= 1
        return unique_triplets

One of LeetCode's solution:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        found, dups = set(), set()
        seen = {}
        for i, val1 in enumerate(nums):
            if val1 not in dups:
                dups.add(val1)
                for j, val2 in enumerate(nums[i+1:]):
                    complement = -val1 - val2
                    if complement in seen and seen[complement] == i:
                        min_val = min((val1, val2, complement))
                        max_val = max((val1, val2, complement))
                        if (min_val, max_val) not in found:
                            found.add((min_val, max_val))
                            res.append([val1, val2, complement])
                    seen[val2] = i
        return res

---

References

• For additional details, you can see the Discussion Board . There are plenty of accepted solutions, explanations, efficient algorithms with a variety of languages, and time/space complexity analysis in there.

Answer 2

Your while loop runs forever if it ever finds a result to put into the results list. That's because the else clause inside your while loop doesn't do anything to change the loop condition, so it will just keep running with the same low , high and i values until you run out of memory to store them.

I suspect you want to advance both low and high after appending a solution to the results:

        while(low<high):
            val = nums[low]+nums[high] + nums[i]
            if(val < 0)  or ((low < high) and nums[low] == nums[low+1]):

                low = low + 1
            elif( val > 0) or ((low < high) and nums[high] == nums[high-1]):

                high = high - 1
            else:
                res.append( (nums[i], nums[low], nums[high]))
                low = low + 1
                high = high - 1