https://leetcode.com/problems/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 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 <= A.length <= 10000-10000 <= A[i] <= 100001 <= queries.length <= 10000-10000 <= queries[i][0] <= 100000 <= queries[i][1] < A.length
- find
sumof all even #s; - for each queries, check the item in A and after-added-up value, if
a) the item in A is even, deduct it fromsum;
b) after-added-up we have an even value, then add the new value tosum;
public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
int S = 0;
for (int x : A)
if (x % 2 == 0)
S += x;
int[] ans = new int[queries.length];
for (int i = 0; i < queries.length; ++i) {
int val = queries[i][0], index = queries[i][1];
if (A[index] % 2 == 0)
S -= A[index];
A[index] += val;
if (A[index] % 2 == 0)
S += A[index];
ans[i] = S;
}
return ans;
}
