查询后的偶数和

/*
 * Problem Statement:
 * Given an integer array A and a query array queries.
 * For the i-th query, let 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 in A.
 * Return an array answer where answer[i] is the answer to the i-th query.
 */

import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {
    public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
        int sumEven = Arrays.stream(A).filter(x -> x % 2 == 0).sum();
        int[] answer = new int[queries.length];
        
        for (int i = 0; i < queries.length; i++) {
            int val = queries[i][0];
            int index = queries[i][1];
            
            if (A[index] % 2 == 0) sumEven -= A[index];
            
            A[index] += val;
            
            if (A[index] % 2 == 0) sumEven += A[index];
            
            answer[i] = sumEven;
        }
        
        return answer;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] A = {1, 2, 3, 4};
        int[][] queries = {{1, 0}, {-3, 1}, {-4, 0}, {2, 3}};
        System.out.println(Arrays.toString(sol.sumEvenAfterQueries(A, queries))); // Output: [8, 6, 2, 4]
    }
}