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
Solution:
class Solution {
public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
int evenSum = 0;
for (int val : A) {
if (val % 2 == 0) {
evenSum += val;
}
}
int n = queries.length;
int[] res = new int[n];
for (int i = 0; i < n; i ++) {
int[] query = queries[i];
int val = A[query[1]];
if (val % 2 == 0) {
evenSum -= val;
}
val += query[0];
if (val % 2 == 0) {
evenSum += val;
}
A[query[1]] = val;
res[i] = evenSum;
}
return res;
}
}