## Algorithm

Problem Name: 985. Sum of Even Numbers After Queries

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, the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums, the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums, the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums, the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
```

Example 2:

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

Constraints:

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

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int[] sumEvenAfterQueries(int[] nums, int[][] queries) {
int evenSum = 0;
for (int num : nums) {
evenSum += num % 2 == 0 ? num : 0;
}
int[] result = new int[nums.length];
for (int i = 0; i < queries.length; i++) {
int idx = queries[i];
if (nums[idx] % 2 == 0) {
evenSum -= nums[idx];
}
nums[idx] += queries[i];
if (nums[idx] % 2 == 0) {
evenSum += nums[idx];
}
result[i] = evenSum;
}
return result;
}
}
``````
Copy The Code &

Input

cmd
nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]

Output

cmd
[8,6,2,4]

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const sumEvenAfterQueries = function(A, queries) {
let sum = A.reduce((acc, cur) => cur%2 == 0 ? acc + cur : acc, 0);
return queries.map((q) => {
let i = q;
let s = A[i] + q;
if(s%2 === 0) {
sum += q;
if(A[i]%2 !== 0) sum += A[i];
} else if(A[i]%2 === 0) sum -= A[i];
A[i] = s;
return sum;
});
};
``````
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Input

cmd
nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]

Output

cmd
[8,6,2,4]

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
sm = sum(a for a in A if a % 2 == 0)
for i in range(len(queries)):
val, ind = queries[i]
sm -= A[ind] % 2 == 0 and A[ind]
A[ind] += val
sm += A[ind] % 2 == 0 and A[ind]
queries[i] = sm
return queries
``````
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Input

cmd
nums = , queries = [[4,0]]

Output

cmd


### #4 Code Example with C# Programming

```Code - C# Programming```

``````
using System.Linq;

namespace LeetCode
{
public class _0985_SumOfEvenNumbersAfterQueries
{
public int[] SumEvenAfterQueries(int[] A, int[][] queries)
{
var sum = A.Where(num => num % 2 == 0).Sum();
var result = new int[queries.Length];

for (int i = 0; i < queries.Length; i++)
{
int val = queries[i], index = queries[i];
if (A[index] % 2 == 0)
sum -= A[index];

A[index] += val;

if (A[index] % 2 == 0)
sum += A[index];

result[i] = sum;
}

return result;
}
}
}
``````
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Input

cmd
nums = , queries = [[4,0]]

Output

cmd