2569. Handling Sum Queries After Update

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        N = len(nums1)
        mask = 1 << N
        nums1 = int("".join(map(str, nums1)), 2)
        total = sum(nums2)
        res = []
        
        for t, i, j in queries:
            if t == 1:
                left = (mask >> i) - 1
                right = (mask >> (j + 1)) - 1
                nums1 = nums1 ^ left ^ right
            elif t == 2:
                total += nums1.bit_count() * i
            else:
                res.append(total)
        
        return res