Watch 6 video solutions for Handling Sum Queries After Update, a hard level problem involving Array, Segment Tree. This walkthrough by Aryan Mittal has 1,681 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:
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.queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.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 <= 105nums1.length = nums2.length1 <= queries.length <= 105queries[i].length = 30 <= l <= r <= nums1.length - 10 <= p <= 1060 <= nums1[i] <= 10 <= nums2[i] <= 109Problem Overview: You maintain two arrays. Queries flip bits in nums1, add a multiple of the number of ones in nums1 to nums2, or ask for the total sum of nums2. The challenge is handling frequent range updates and keeping the count of ones available instantly for later queries.
Approach 1: Naive Iterative Approach (O(n) per update, O(q·n) overall)
The straightforward method directly applies each query. For a flip operation, iterate through the specified range in nums1 and toggle each bit. For the second query type, count the number of ones in nums1 by iterating through the array, then add p * ones to a running total of nums2. This approach relies purely on array traversal using basic operations on an array. It works for small inputs but becomes too slow when both the array size and number of queries grow large.
Approach 2: Segment Tree Optimization for Range Updates (O((n + q) log n) time, O(n) space)
The optimized approach stores the count of ones in nums1 using a segment tree. Each node tracks how many ones exist in its range. When a query flips a range, instead of updating each element individually, the tree performs a lazy range flip: the stored count becomes segment_length - current_ones. Lazy propagation postpones updates to children until necessary, keeping operations efficient.
For the second query type, the algorithm reads the total number of ones directly from the root of the tree in O(1). Multiply that value by p and add it to a running sum representing the total of nums2. Since the problem only asks for the sum of nums2, there is no need to update each element individually. The third query simply outputs this stored sum. This design converts expensive full-array operations into logarithmic updates and constant-time queries.
Recommended for interviews: Interviewers expect the Segment Tree with lazy propagation solution. The brute-force iteration shows you understand the mechanics of the queries, but it does not scale. Using a segment tree to maintain the number of ones while supporting fast range flips demonstrates strong knowledge of range query data structures.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Naive Iterative Approach | O(q · n) | O(1) | Small arrays or when implementing a quick baseline solution |
| Segment Tree with Lazy Propagation | O((n + q) log n) | O(n) | Large input sizes with many range flips and queries |