Question

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

Difficulty:Easy

Category:

Analyze

Solution

class Solution {
 public:
  vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& q) {
    int n = q.size();
    vector<int> ans(n, 0);
    for (int i = 0; i < n; ++i) {
      int val = q[i][0], idx = q[i][1];
      A[idx] += val;
      for (int j = 0; j < n; ++j) {
        if (A[j] & 1) {
          continue;
        } else {
          ans[i] += A[j];
        }
      }
    }
    return ans;
  }
};