985. Sum of Even Numbers After Queries

1. Description

You are given an integer array nums and an array queries where queries[i] = [val$_i$, index$_i$].
For each query i, first, apply nums[index$_i$] = nums[index$_i$] + val$_i$, then print the sum of the even values of nums.
Return an integer array answer where answer[i] is the answer to the i$^{th}$ query.

2. Example

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]

3. Constraints

• 1 <= nums.length <= 10$^4$
• -10$^4$ <= nums[i] <= 10$^4$
• 1 <= queries.length <= 10$^4$
• -10$^4$ <= val$_i$ <= 10$^4$
• 0 <= index$_i$ < nums.length

4. Solutions

Simulation

n = nums.size(), q = queries.size()
Time complexity: O(n + q)
Space complexity: O(1)

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class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int> nums, const vector<vector<int>> &queries) {
        int sum = accumulate(nums.begin(), nums.end(), 0, [](int sum, int item) {
            return (item & 1) == 0 ? sum + item : sum;
        });

        vector<int> sums;
        sums.reserve(queries.size());

        for (const auto &query : queries) {
            int index = query[1];
            int value = query[0];

            int origin_value = nums[index];
            nums[index] += value;

            if ((origin_value & 1) == 0) {
                sum -= origin_value;
            }

            if ((nums[index] & 1) == 0) {
                sum += nums[index];
            }

            sums.push_back(sum);
        }

        return sums;
    }
};