985. Sum of Even Numbers After Queries

Problem

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. 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

Problem Summary

Given an integer array A and a query array queries.

For the i-th query, there are 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 in A. (Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.) Return the answers to all queries. Your answer should be given as an array answer, where answer[i] is the answer to the i-th query.

Solution Approach

• Given an array A and a query array. Each query operation is required to change the value of an element in array A and compute the sum of the even values in array A after this operation.
• Easy problem. First compute the sum of all even numbers in A. Then, during each query operation, dynamically maintain this sum of even numbers.

Code

package leetcode

func sumEvenAfterQueries(A []int, queries [][]int) []int {
	cur, res := 0, []int{}
	for _, v := range A {
		if v%2 == 0 {
			cur += v
		}
	}
	for _, q := range queries {
		if A[q[1]]%2 == 0 {
			cur -= A[q[1]]
		}
		A[q[1]] += q[0]
		if A[q[1]]%2 == 0 {
			cur += A[q[1]]
		}
		res = append(res, cur)
	}
	return res
}