2569 - Handling Sum Queries After Update\.

Hard

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:

• <code>1 <= nums1.length,nums2.length <= 10⁵</code>

• nums1.length = nums2.length

• <code>1 <= queries.length <= 10⁵</code>

• queries[i].length = 3

• 0 <= l <= r <= nums1.length - 1

• <code>0 <= p <= 10⁶</code>

• 0 <= nums1[i] <= 1

• <code>0 <= nums2i<= 10⁹</code>