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[9/21] 985. Sum of Even Numbers After Queries
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Medium
1784308Add to ListShareYou 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] <= 1041 <= queries.length <= 104-104 <= vali <= 1040 <= indexi < nums.length
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Submissions
178,428
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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.
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
