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:

Input: nums = [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 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.

Example 2:

Input: nums = [1], queries = [[4,0]]
Output:
 [0]

Constraints:

• 1 <= nums.length <= 104

• -104 <= nums[i] <= 104

• 1 <= queries.length <= 104

• -104 <= vali <= 104

• 0 <= indexi < nums.length

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