## Problem

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

For the i-th query val = queries[i], index = queries[i], 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] 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, the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A, the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A, the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A, the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.


Note:

1. 1 <= A.length <= 10000
2. -10000 <= A[i] <= 10000
3. 1 <= queries.length <= 10000
4. -10000 <= queries[i] <= 10000
5. 0 <= queries[i] < A.length

## Solution: Simulation

Time complexity: O(n + |Q|)
Space complexity: O(n)

## Python3

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