{ "data": { "question": { "questionId": "2703", "questionFrontendId": "2569", "boundTopicId": null, "title": "Handling Sum Queries After Update", "titleSlug": "handling-sum-queries-after-update", "content": "
You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:
queries[i] = [1, l, r]. Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r. Both l and r are 0-indexed.queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.queries[i] = [3, 0, 0]. Find the sum of the elements in nums2.Return an array containing all the answers to the third type queries.
\n\n\n
Example 1:
\n\n\nInput: nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]\nOutput: [3]\nExplanation: 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.\n\n\n
Example 2:
\n\n\nInput: nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]\nOutput: [5]\nExplanation: After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.\n\n\n
\n
Constraints:
\n\n1 <= nums1.length,nums2.length <= 105nums1.length = nums2.length1 <= queries.length <= 105queries[i].length = 30 <= l <= r <= nums1.length - 10 <= p <= 1060 <= nums1[i] <= 10 <= nums2[i] <= 109Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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Python 2.7.12.
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Some common data structure implementations are provided in the Algorithms module: https://www.rubydoc.info/github/kanwei/algorithms/Algorithms
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