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.

 

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.
'''

class Solution:
    def sumEvenAfterQueries(self, A: 'List[int]', queries: 'List[List[int]]') -> 'List[int]':
        sums = list()
        sum_of_numbers = sum(a for a in A if a%2==0)
                
        for item in queries:
            t = A[item[1]]
            A[item[1]] += item[0]
            if A[item[1]] % 2 == 0:
                if t % 2 == 0:
                    sum_of_numbers += - t + A[item[1]]
                else:
                    sum_of_numbers += A[item[1]]
            else:
                if t % 2 == 0:
                    sum_of_numbers = sum_of_numbers - t
            sums.append(sum_of_numbers)
            
        return sums