## Solution

#### Approach 1: Maintain Array Sum

Intuition and Algorithm

Let's try to maintain S, the sum of the array throughout one query operation.

When acting on an array element A[index], the rest of the values of A remain the same. Let's remove A[index] from S if it is even, then add A[index] + val back (if it is even.)

Here are some examples:

• If we have A = [2,2,2,2,2], S = 10, and we do A[0] += 4: we will update S -= 2, then S += 6. At the end, we will have A = [6,2,2,2,2] and S = 14.

• If we have A = [1,2,2,2,2], S = 8, and we do A[0] += 3: we will skip updating S (since A[0] is odd), then S += 4. At the end, we will have A = [4,2,2,2,2] and S = 12.

• If we have A = [2,2,2,2,2], S = 10 and we do A[0] += 1: we will update S -= 2, then skip updating S (since A[0] + 1 is odd.) At the end, we will have A = [3,2,2,2,2] and S = 8.

• If we have A = [1,2,2,2,2], S = 8 and we do A[0] += 2: we will skip updating S (since A[0] is odd), then skip updating S again (since A[0] + 2 is odd.) At the end, we will have A = [3,2,2,2,2] and S = 8.

These examples help illustrate that our algorithm actually maintains the value of S throughout each query operation.

Complexity Analysis

• Time Complexity: , where is the length of A and is the number of queries.

• Space Complexity: , though we only allocate additional space.

Analysis written by: @awice.