'''
## Questions
### 985. [Sum of Even Numbers After Queries](https://leetcode.com/problems/sum-of-even-numbers-after-queries/)
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
'''
## Solutions

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

# Runtime: 608 ms
# Memory Usage: 19.7 MB
