985. Sum of Even Numbers After Queries-查询后的偶数和

We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.

Example 1:
Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation:
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Note:

• 1 <= A.length <= 10000
• -10000 <= A[i] <= 10000
• 1 <= queries.length <= 10000
• -10000 <= queries[i][0] <= 10000
• 0 <= queries[i][1] < A.length

Solution in python:

class Solution:
    def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
        result = []
        total = sum(i for i in A if i % 2 ==0)
        for item in queries:
            if A[item[1]] % 2 == 1 and item[0] % 2 == 1:
                total += A[item[1]] + item[0]
            elif A[item[1]] % 2 == 0 and item[0] % 2 == 1:
                total -= A[item[1]]
            elif A[item[1]] % 2 == 0 and item[0] % 2 == 0:
                total += item[0]
            A[item[1]] += item[0]
            result.append(total)
        return result