[LeetCode]1013. Partition Array Into Three Parts With Equal Sum

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])

Example 1:

Input: A = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Example 2:

Input: A = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false
Example 3:

Input: A = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
 

Constraints:

3 <= A.length <= 50000
-10^4 <= A[i] <= 10^4

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/partition-array-into-three-parts-with-equal-sum
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python3:

class Solution:
    def canThreePartsEqualSum(self, A: List[int]) -> bool:
        if sum(A) % 3 != 0:
            return False
        
        avg = sum(A) // 3
        i = 0
        j = len(A) - 1
        sum_f = A[i]
        i += 1
        sum_b = A[j]
        j -= 1
        rst = False
        
        while i < j:
            if sum_f != avg:
                sum_f += A[i]
                i += 1
            if sum_b != avg:
                sum_b += A[j]
                j -= 1
            if sum_f == avg and sum_b == avg:
                rst = True
                break
        
        if i > j:
            rst = False
        return rst