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.
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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.
Java Solution
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class Solution {
public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
int N = A.length;
int[] sumEventQueries = new int[N];
int num = queries.length;
for(int i = 0; i < num; i++){
int index = queries[i][1];
int val = queries[i][0];
A[index] = A[index] + val;
for(int ele: A){
if(isEvent(ele)){
sumEventQueries[i] += ele;
}
}
}
return sumEventQueries;
}
private boolean isEvent(int a){
if(a % 2 == 0){
return true;
}
else
return false;
}
}
