985. Sum of Even Numbers After Queries

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. 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

Rust Solution

struct 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);
}

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