Leetcode daily question (985. sum of even numbers after queries)

You are given an integer array nums and an array queries where queries[i] = [vali, indexi].

For each query i, first, apply nums[indexi] = nums[indexi] + vali, then print the sum of the even values of nums.

Return an integer array answer where answer[i] is the answer to the ith query.

Example 1:

Input: nums = [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 nums[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Example 2:

Input: nums = [1], queries = [[4,0]]
Output: [0]

Constraints:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• 1 <= queries.length <= 104
• -104 <= vali <= 104
• 0 <= indexi < nums.length

---

consider 4 In this case :

1. nums[i] It turned out to be an even number , add val Then even numbers , So the even sum increases val
2. nums[i] It turned out to be an even number , add val Then it becomes odd , In this way, the even sum is reduced nums[i]
3. nums[i] It turned out to be an odd number , add val Then it becomes even , So the even sum increases nums[i] + val
4. nums[i] It turned out to be an odd number , add val Then there are odd numbers , So there is no change in the even sum

---

impl Solution {
    
    pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
    
        let mut sum = nums.iter().filter(|v| *v % 2 == 0).map(|v| *v).sum();
        let mut ans = Vec::new();
        for q in queries {
    
            let curr = nums[q[1] as usize];
            if curr % 2 == 0 {
    
                if (curr + q[0]) % 2 == 0 {
    
                    sum += q[0];
                } else {
    
                    sum -= curr;
                }
            } else {
    
                if (curr + q[0]) % 2 == 0 {
    
                    sum += curr + q[0];
                }
            }
            nums[q[1] as usize] += q[0];
            ans.push(sum);
        }
        ans
    }
}