2569. Handling Sum Queries After Update
Hard
123
21

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`
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