Question 523 Continuous SubArray Sum

The description of the code is in the following link

 

My Code

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        if len(nums) <2:
            return False
        else:
            if k == 0:
                InputList= nums[0:2]
                if sum(InputList)  == 0:
                    return True
                else:
                    SumList = [[sum(InputList[i:len(InputList)]) for i in range(len(InputList))]]
                    SumList = [[0],[0]]+SumList # To get the save for the list with length 2
                    for i in range(2, len(nums)):
                        NewList = []
                        for j in range(i):
                            NewSum = nums[i]+SumList[i][j]
                            if NewSum == 0:
                                return True
                            NewList.append(NewSum)
                        NewList.append(nums[i])
                        SumList.append(NewList)
                    return False
            else:
                InputList = nums[0:2];
                if sum(InputList) % k ==0:
                    return True
                else:
                    SumList = [[sum(InputList[i:len(InputList)]) for i in range(len(InputList))]]
                    SumList = [[0],[0]]+SumList # To get the save for the list with length 2
                    for i in range(2, len(nums)):
                        NewList = []
                        for j in range(i):
                            NewSum = nums[i]+SumList[i][j]
                            if NewSum %k  == 0:
                                return True
                            NewList.append(NewSum)
                        NewList.append(nums[i])
                        SumList.append(NewList)
                return False

Explanation: The key point is still doing it brutally. We check all the possible sum and whenever we encounter find there is a full division, we return true, otherwise we return false. Here we use one extra array to improve the time complexity, otherwise it is going to be n^3, now it is n^2

 

Another Approach: Using Hash Table (The Solution is not clearly clarified, need to do some further reading)