Watch 6 video solutions for XOR After Range Multiplication Queries I, a medium level problem involving Array, Divide and Conquer, Simulation. This walkthrough by Eduardo Nakanishi has 230 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer array nums of length n and a 2D integer array queries of size q, where queries[i] = [li, ri, ki, vi].
For each query, you must apply the following operations in order:
idx = li.idx <= ri:
nums[idx] = (nums[idx] * vi) % (109 + 7)idx += ki.Return the bitwise XOR of all elements in nums after processing all queries.
Example 1:
Input: nums = [1,1,1], queries = [[0,2,1,4]]
Output: 4
Explanation:
[0, 2, 1, 4] multiplies every element from index 0 through index 2 by 4.[1, 1, 1] to [4, 4, 4].4 ^ 4 ^ 4 = 4.Example 2:
Input: nums = [2,3,1,5,4], queries = [[1,4,2,3],[0,2,1,2]]
Output: 31
Explanation:
[1, 4, 2, 3] multiplies the elements at indices 1 and 3 by 3, transforming the array to [2, 9, 1, 15, 4].[0, 2, 1, 2] multiplies the elements at indices 0, 1, and 2 by 2, resulting in [4, 18, 2, 15, 4].4 ^ 18 ^ 2 ^ 15 ^ 4 = 31.
Constraints:
1 <= n == nums.length <= 1031 <= nums[i] <= 1091 <= q == queries.length <= 103queries[i] = [li, ri, ki, vi]0 <= li <= ri < n1 <= ki <= n1 <= vi <= 105Problem Overview: You are given an array and a list of queries. Each query multiplies every element in a subarray [l, r] by a value. After applying the operations, compute the XOR of the resulting array. The main challenge is handling repeated range updates while keeping the final XOR correct.
Approach 1: Direct Simulation (O(n * q) time, O(1) space)
The most straightforward strategy is to simulate every query exactly as described. For each query, iterate from l to r and multiply the element by the given value. After processing all queries, iterate once through the array and compute the XOR of all elements. This approach uses plain array traversal and mirrors the problem statement directly, which makes it easy to implement and debug.
The key observation is that multiplication changes multiple bits in unpredictable ways, so maintaining a running XOR during updates becomes tricky. Instead of attempting clever bit manipulations, simply update the array values and compute the XOR at the end. For typical constraints in this problem, the straightforward simulation is efficient enough.
Approach 2: Incremental XOR Update (O(n * q) time, O(1) space)
A small optimization is to maintain the global XOR while applying updates. When an element a[i] changes, remove its previous contribution from the XOR using xor ^= oldValue, update the element with the multiplication, then add the new value back using xor ^= newValue. This avoids recomputing the XOR over the entire array at the end.
The update loop for each query still iterates from l to r, so the overall complexity remains the same. However, the final XOR is always available without an additional pass. This pattern often appears in problems combining range updates with XOR aggregation and simple simulation.
Recommended for interviews: Start with the direct simulation approach. It demonstrates that you understand the mechanics of the queries and how XOR aggregation works. If the interviewer asks for improvements, explain how maintaining the XOR incrementally avoids an extra pass. Problems tagged with divide and conquer sometimes admit segment-tree optimizations, but for this version the straightforward simulation is usually expected.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Direct Simulation | O(n * q) | O(1) | Best when constraints are small or moderate and implementation speed matters |
| Simulation with Incremental XOR Maintenance | O(n * q) | O(1) | Useful when you want the XOR updated during each modification without an additional array pass |
| Segment Tree / Divide and Conquer Idea | O((n + q) log n) | O(n) | Consider when constraints are very large and frequent range updates require structured queries |