# 985. Sum of Even Numbers After Queries

#### Medium

You are given an integer array `nums` and an array `queries` where `queries[i] = [vali, indexi]`.

For each query `i`, first, apply `nums[indexi] = nums[indexi] + vali`, then print the sum of the even values of `nums`.

Return *an integer array* `answer` *where* `answer[i]` *is the answer to the* `ith` *query*.

 

**Example 1:**

<pre><code>Input: nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
<strong>Output:
</strong> [8,6,2,4]
<strong>Explanation:
</strong> At the beginning, the array is [1,2,3,4].
After adding 1 to nums[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
</code></pre>

**Example 2:**

<pre><code>Input: nums = [1], queries = [[4,0]]
<strong>Output:
</strong> [0]
</code></pre>

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**Constraints:**

* `1 <= nums.length <= 104`
* `-104 <= nums[i] <= 104`
* `1 <= queries.length <= 104`
* `-104 <= vali <= 104`
* `0 <= indexi < nums.length`

```python
class Solution:
    def sumEvenAfterQueries(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        ans, curr = [], sum(n for n in nums if n % 2 == 0)
        for val, index in queries:
            prev = nums[index]
            nums[index] += val
            if prev % 2 == 0:
                curr -= prev
            if nums[index] % 2 == 0:
                curr += nums[index]
            ans.append(curr)
        return ans
```