985-sum-of-even-numbers-after-queries.py

"""
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.
Note:

    1. 1 <= A.length <= 10000
    2. -10000 <= A[i] <= 10000
    3. 1 <= queries.length <= 10000
    4. -10000 <= queries[i][0] <= 10000
    5. 0 <= queries[i][1] < A.length

Example:

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.
"""


class Solution(object):
    def sumEvenAfterQueries(self, A, queries):
        """
        :type A: List[int]
        :type queries: List[List[int]]
        :rtype: List[int]
        """
        s = 0
        ans = []
        s = sum(num for num in A if num % 2 == 0)
        for val, idx in queries:
            if A[idx] % 2 == 0:
                s -= A[idx]
            A[idx] += val
            if A[idx] % 2 == 0:
                s += A[idx]
            ans.append(s)
        return ans


if __name__ == "__main__":
    A = [1, 2, 3, 4]
    queries = [[1, 0], [-3, 1], [-4, 0], [2, 3]]
    print(Solution().sumEvenAfterQueries(A, queries))