The simple problem of leetcode is to divide an array into three parts equal to sum

subject

Give you an array of integers arr, Only can it be divided into three and equal Non empty Return only when partial true, Otherwise return to false.
Formally , If you can find the index i + 1 < j And meet (arr[0] + arr[1] + … + arr[i] == arr[i + 1] + arr[i + 2] + … + arr[j - 1] == arr[j] + arr[j + 1] + … + arr[arr.length - 1]) You can divide the array into three equal parts .
Example 1：
Input ：arr = [0,2,1,-6,6,-7,9,1,2,0,1]
Output ：true
explain ：0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Example 2：
Input ：arr = [0,2,1,-6,6,7,9,-1,2,0,1]
Output ：false
Example 3：
Input ：arr = [3,3,6,5,-2,2,5,1,-9,4]
Output ：true
explain ：3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
Tips ：
3 <= arr.length <= 5 * 10^4
-10^4 <= arr[i] <= 10 ^4
source ： Power button （LeetCode）

Their thinking

   The sum of the three parts is equal , Then the sum of one part is the sum of one third . Sum of traversal array , When there is a situation equal to one third of the sum in the summation process , Then the currently traversed elements can form one or two groups . in addition , Consider the special case of sum of arrays ; Assume that all elements of the array are summed to 0, Then the array may be divided into n Share , Just to prove ：sum/3×n=sum. If n≠3 that sum=0.

class Solution:
    def canThreePartsEqualSum(self, arr: List[int]) -> bool:
        total=sum(arr)
        part=total/3
        s=0
        count=0
        for i in arr:
            s+=i
            if s==part:
                count+=1
                s=0
        return True if count>=3 else False # Suppose there are four parts , Each part is the sum of one-third , Then the sum of the current array must be 0