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] <= 100001 <= queries.length <= 10000-10000 <= queries[i][0] <= 100000 <= queries[i][1] < A.lengthstruct Solution;
impl Solution {
fn sum_even_after_queries(mut a: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
let mut sum = a.iter().filter(|&x| x % 2 == 0).sum();
let mut res: Vec<i32> = vec![];
for query in queries {
let v = query[0];
let i = query[1] as usize;
let x = a[i];
let y = a[i] + v;
if x % 2 == 0 {
sum -= x;
}
if y % 2 == 0 {
sum += y;
}
a[i] = y;
res.push(sum);
}
res
}
}
#[test]
fn test() {
let a = vec![1, 2, 3, 4];
let queries: Vec<Vec<i32>> = vec_vec_i32![[1, 0], [-3, 1], [-4, 0], [2, 3]];
let res = vec![8, 6, 2, 4];
assert_eq!(Solution::sum_even_after_queries(a, queries), res);
}