Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 21

Description

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 <= 10^4
• -104 <= nums[i] <= 10^4
• 1 <= queries.length <= 10^4
• -104 <= vali <= 10^4
• 0 <= indexi < nums.length

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<int> sumEvenAfterQueries(vector<int>& nums, vector<vector<int>>& queries) {
    vector<int> answer;
    answer.reserve(queries.size());

    int current = accumulate(nums.begin(), nums.end(), 0, [](int s, int a) {
      return s + !(a & 1) * a;
    });
    for(const auto &query : queries) {
      if(!(nums[query[1]] & 1)) {
        current -= nums[query[1]];
      }
      nums[query[1]] += query[0];
      if(!(nums[query[1]] & 1)) {
        current += nums[query[1]];
      }
      answer.push_back(current);
    }
    return answer;
  }
};

// Accepted
// 58/58 cases passed (191 ms)
// Your runtime beats 49.8 % of cpp submissions
// Your memory usage beats 85.52 % of cpp submissions (45.4 MB)