#2569

Handling Sum Queries After Update

international master · 1975 · lc hard +32 · verified · 30.5% accepted · 198 likes · top 8%

Description

You are given binary arrays nums1 and nums2 and a query list with three types: type 1 [1, l, r] flips all bits in nums1[l..r]; type 2 [2, p, 0] adds nums1[i] * p to each nums2[i]; type 3 [3, 0, 0] requests the current sum of nums2. Return the answers to all type-3 queries in order.

Example 1:

Input: nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]

Output: [3]

Explanation: After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.

Example 2:

Input: nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]

Output: [5]

Explanation: After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.

Code

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class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:

class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]: