Sum Of Even Numbers After Queries Problem
Description
LeetCode Problem 985.
You are given an integer array nums and an array queries where queries[i] = [val_i, index_i].
For each query i, first, apply nums[index_i] = nums[index_i] + val_i, then print the sum of the even values of nums.
Return an integer array answer where answer[i] is the answer to the i^th query.
Example 1:
1
2
3
4
5
6
7
Input: nums = [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 nums[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
Example 2:
1
2
Input: nums = [1], queries = [[4,0]]
Output: [0]
Constraints:
- 1 <= nums.length <= 10^4
- -10^4 <= nums[i] <= 10^4
- 1 <= queries.length <= 10^4
- -10^4 <= val_i <= 10^4
- 0 <= index_i < nums.length
Sample C++ Code
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public:
vector<int> sumEvenAfterQueries(vector<int>& nums, vector<vector<int>>& queries) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] % 2 == 0) {
sum += nums[i];
}
}
vector<int> ans;
for (int i = 0; i < queries.size(); i++) {
int val = queries[i][0];
int index = queries[i][1];
if (nums[index] % 2 == 0) {
sum -= nums[index];
}
nums[index] += val;
if (nums[index] % 2 == 0) {
sum += nums[index];
}
ans.push_back(sum);
}
return ans;
}
};