Q985 Sum of Even Numbers After Queries

Sum of Even Numbers After Queries

Question

We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index]. Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries. Your answer array should have answer[i] as the answer to the i-th query.

Example 1:

Input: A = [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 A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8. After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6. After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2. After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4

Note:

• 1 <= A.length <= 10000
• -10000 <= A[i] <= 10000
• 1 <= queries.length <= 10000
• -10000 <= queries[i][0] <= 10000
• 0 <= queries[i][1] < A.length

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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.

class Solution:
    def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
        results=[]
        even = sum(x for x in A if x % 2 == 0)
        for x,y in queries:
            if x%2!=0 and A[y]%2!=0:
                even+=(x+A[y])
            elif x%2==0 and A[y]%2==0:
                even += x
            elif x % 2 != 0 and A[y] % 2 == 0:
                even -= A[y]
            A[y] += x
            results.append(even)
        return results

• Runtime: 180 ms, faster than 49.38% of Python3 online submissions forSum of Even Numbers After Queries.
• Memory Usage: 17.5 MB, less than 5.56% of Python3 online submissions for Sum of Even Numbers After Queries.