g2501_2600.s2569_handling_sum_queries_after_update.readme.md Maven / Gradle / Ivy
Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation
Show all versions of leetcode-in-java21 Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
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:
1. 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**.
2. 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`.
3. 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