# Leetcode: Sum of Even Numbers After Queries

Sum of Even Numbers After Queries

Similar Problems:

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.)

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

// Basic Ideas: array
// Complexity: Time O(n+k), Space O(1)
func abs(num int) int {
if num < 0 { num = -num }
return num
}
func sumEvenAfterQueries(A []int, queries [][]int) []int {
even_sum := 0
res := []int{}
for _, num := range A {
if num%2 == 0 { even_sum += num }
}

for _, q := range queries {
val, index := q[0], q[1]
if abs(A[index] + val)%2 == 0 {
if abs(A[index])%2 == 0 {
even_sum += val
} else {
even_sum += val+A[index]
}
}
if abs(A[index])%2 == 0 && abs(val)%2 == 1 {
even_sum -= A[index]
}
A[index] += val
res = append(res, even_sum)
}
return res
}

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