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.

 

 

class Solution {
public:
  vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
    int initval = 0;
    vector<int> ret;
    for(int v : A) {
      if(v % 2 == 0) initval += v;
    }
    for(auto v : queries) {
      int before = A[v[1]];
      A[v[1]] += v[0];
      if(before % 2 == 0) initval -= before;
      if(A[v[1]] % 2 == 0) initval += A[v[1]];
      ret.push_back(initval);
    }
    return ret;
  }
};

 

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