985

985. Sum of Even Numbers After Queries (Easy)

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

Companies:
Indeed

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/sum-of-even-numbers-after-queries/
// Author: github.com/lzl124631x
// Time: O(A + Q)
// Space: O(1)
class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
        vector<int> ans(queries.size());
        int sum = 0;
        for (int n : A) {
            if (n % 2 == 0) sum += n;
        }
        for (int i = 0; i < queries.size(); ++i) {
            auto q = queries[i];
            int val = q[0], index = q[1], prev = A[index];
            A[index] += val;
            bool wasEven = prev % 2 == 0, even = A[index] % 2 == 0;
            if (wasEven && even) {
                ans[i] = (sum += val);
            } else if (wasEven && !even) {
                ans[i] = (sum -= prev);
            } else if (!wasEven && even) {
                ans[i] = (sum += A[index]);
            } else ans[i] = sum;
        }
        return ans;
    }
};