2569. Handling Sum Queries After Update ¶

Problem

You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:

For a query of type 1, 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.
For a query of type 2, queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.
For a query of type 3, 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.

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.

Constraints:

1 <= nums1.length,nums2.length <= 10⁵
nums1.length = nums2.length
1 <= queries.length <= 10⁵
queries[i].length = 3
0 <= l <= r <= nums1.length - 1
0 <= p <= 10⁶
0 <= nums1[i] <= 1
0 <= nums2[i] <= 10⁹