• Home
• com.github.javadev
• leetcode-in-java
• 1.38
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g2501_2600.s2569_handling_sum_queries_after_update.readme.md Maven / Gradle / Ivy

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