985. Sum of Even Numbers After Queries

Problem

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 <= 10^4

• -10^4 <= nums[i] <= 10^4

• 1 <= queries.length <= 10^4

• -10^4 <= vali <= 10^4

• 0 <= indexi < nums.length

Solution (Java)

class Solution {
    public int[] sumEvenAfterQueries(int[] nums, int[][] queries) {
        int[] result = new int[queries.length];
        int res = 0;
        for (int num : nums) {
            res += (num & 1) == 0 ? num : 0;
        }
        int k = 0;
        for (int[] query : queries) {
            res -= (nums[query[1]] & 1) == 0 ? nums[query[1]] : 0;
            nums[query[1]] += query[0];
            if ((nums[query[1]] & 1) == 0) {
                res += nums[query[1]];
            }
            result[k++] = res;
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).