[leetcode] problem 985 – sum of even numbers after queries

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

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 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length

Code

public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
    int n = queries.length;
    int[] result = new int[n];
    int sum = 0;

    for (int i = 0; i < A.length; i++) {
        if (A[i] % 2 == 0)
            sum += A[i];
    }

    for (int i = 0; i < n; i++) {
        int val = queries[i][0];
        int index = queries[i][1];

        if (A[index] % 2 == 0)
            sum -= A[index];

        A[index] += val;

        if (A[index] % 2 == 0)
            sum += A[index];

        result[i] = sum;
    }

    return result;
}

[leetcode] problem 985 – sum of even numbers after queries

Example

Note

Code

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