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# [9/21] 985. Sum of Even Numbers After Queries

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### 본문

Medium

You are given an integer array `nums` and an array `queries` where `queries[i] = [vali, indexi]`.

For each query `i`, first, apply `nums[indexi] = nums[indexi] + vali`, then print the sum of the even values of `nums`.

Return an integer array `answer` where `answer[i]` is the answer to the `ith` query.

Example 1:

```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:

```Input: nums = [1], queries = [[4,0]]
Output: [0]
```

Constraints:

• `1 <= nums.length <= 104`
• `-104 <= nums[i] <= 104`
• `1 <= queries.length <= 104`
• `-104 <= vali <= 104`
• `0 <= indexi < nums.length`
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## 학부유학생님의 댓글

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Runtime: 540 ms, faster than 90.84% of Python3 online submissions for Sum of Even Numbers After Queries.
Memory Usage: 20.5 MB, less than 44.27% of Python3 online submissions for Sum of Even Numbers After Queries.
``````class Solution:
def sumEvenAfterQueries(self, nums: List[int], queries: List[List[int]]) -> List[int]:
evensum = sum(num for num in nums if not num%2)
res = []
for val, idx in queries:
new_val = nums[idx] + val
#odd to even
if nums[idx]%2 and new_val%2 == 0:
evensum += new_val
# even to odd
elif nums[idx]%2 == 0 and new_val%2:
evensum -= nums[idx]

# even to even
elif nums[idx]%2 == 0 and new_val%2 == 0:
evensum += new_val - nums[idx]

nums[idx] = new_val

res.append(evensum)
# print(evensum, nums)
return res
``````
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