985 Sum of Even Numbers After Queries.py

#!/usr/bin/python3
"""
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
"""
from typing import List


class Solution:
    def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
        """
        maintain a sum
        """
        cur_sum = sum(filter(lambda x: x % 2 == 0, A))
        ret = []
        for val, idx in queries:
            prev = A[idx]
            if prev % 2 == 0:
                cur_sum -= prev
            A[idx] += val
            if A[idx] % 2 == 0:
                cur_sum += A[idx]
            ret.append(cur_sum)

        return ret