sumEvenAfterQueries.go

/* https://leetcode.com/problems/sum-of-even-numbers-after-queries/description/
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
*/
package larray

func sumEvenAfterQueries(A []int, queries [][]int) []int {
	tmp := make([]int, len(A), len(A))
	copy(tmp, A)

	sum := 0
	for _, value := range tmp {
		if value%2 == 0 {
			sum += value
		}
	}

	res := make([]int, len(A), len(A))
	for i, query := range queries {
		value, idx := query[0], query[1]
		old := tmp[idx]
		tmp[idx] += value
		if tmp[idx]%2 == 0 {
			if old%2 == 0 {
				sum += value
			} else {
				sum += tmp[idx]
			}
		} else {
			if old%2 == 0 {
				sum = sum - old
			}
		}
		res[i] = sum
	}
	return res
}