[Solution] 985. Sum of Even Numbers After Queries

Posted 2021-02-07 downstream-1998

• Difficulty: Easy

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 the 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

Related Topics

Array

Solution

1. 求出原数组中的偶数和

2. 对于每一个查询，对于数组元素 A[queries[i][1]] 的影响，按以下四种情况处理：

   • 之前为奇数，之后为奇数：无需操作；
   • 之前为奇数，之后为偶数：在原偶数和的基础上加上这个新增的偶数；
   • 之前为偶数，之后为奇数：在原偶数和的基础上去掉这个之前的偶数；
   • 之前为偶数，之后为奇数：在原偶数和的基础上加上一个变化量（可能为正，也可能为负）；

   将每次查询得到的偶数和放入结果数组中即为所求。

public class Solution
{
    public int[] SumEvenAfterQueries(int[] A, int[][] queries)
    {
        int[] ret = new int[queries.GetLength(0)];
        int sum = (from x in A where x % 2 == 0 select x).Sum();

        for(int i = 0; i < queries.GetLength(0); i++)
        {
            int before = A[queries[i][1]];
            A[queries[i][1]] += queries[i][0];
            if(before % 2 == 0)
            {
                if(A[queries[i][1]] % 2 == 0)
                {
                    int delta = A[queries[i][1]] - before;
                    sum += delta;
                }
                else
                {
                    sum -= before;
                }
            }
            else
            {
                if(A[queries[i][1]] % 2 == 0)
                {
                    sum += A[queries[i][1]];
                }
                else
                {
                    // no operation
                }
            }
            ret[i] = sum;
        }

        return ret;
    }
}