# Solving Leetcode Interviews in Seconds with AI: Handling Sum Queries After Update


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2569" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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:  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  

	# Explanation
	Here's an efficient solution to the problem, incorporating the constraints and optimization considerations:

*   **Differential Array for Flips:** Use a differential array to efficiently handle the range updates (flips) in `nums1`. This avoids iterating through the entire range for each flip operation. The differential array stores the difference between consecutive elements, allowing us to perform range updates in O(1) time.
*   **Direct Updates to nums2:** Directly update `nums2` based on the values in `nums1` for type 2 queries.
*   **Sum Calculation:** Calculate the sum of `nums2` when type 3 queries are encountered and store the results.

*   **Runtime Complexity:** O(Q + N), where Q is the number of queries and N is the length of the arrays.
*   **Storage Complexity:** O(N), due to the differential array.

	
	# Code
	```python
	def handle_queries(nums1, nums2, queries):
    n = len(nums1)
    diff = [0] * (n + 1)  # Differential array for nums1

    # Initialize differential array
    for i in range(n):
        diff[i] = nums1[i]
        if i > 0:
            diff[i] -= nums1[i-1]
    diff[n] = -nums1[n-1]

    results = []
    for query in queries:
        query_type = query[0]

        if query_type == 1:
            l, r = query[1], query[2]
            # Flip range [l, r] in nums1 using the differential array
            if diff[l] == 0:
                diff[l] = 1
            else:
                diff[l] = 0
            if r + 1 <= n - 1:
                if diff[r + 1] == 0:
                    diff[r + 1] = 1
                else:
                    diff[r + 1] = 0
            elif r + 1 == n:
                if diff[r + 1] == 0:
                    diff[r + 1] = 1
                else:
                    diff[r + 1] = 0


            #update nums1 based on diff array
            current_val = 0
            for i in range(n):
                if i == 0:
                    nums1[i] = diff[i]
                else:
                    nums1[i] = nums1[i-1] + diff[i]
                nums1[i] = nums1[i]%2
        elif query_type == 2:
            p = query[1]
            # Update nums2
            for i in range(n):
                nums2[i] += nums1[i] * p
        elif query_type == 3:
            # Calculate the sum of nums2
            results.append(sum(nums2))

    return results
	```