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.)

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

Github: code.dennyzhang.com

Credits To: leetcode.com

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

// Blog link: https://code.dennyzhang.com/sum-of-even-numbers-after-queries // 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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