How do you solve XOR After Range Multiplication Queries II on LeetCode?

XOR After Range Multiplication Queries II (LeetCode #3655) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Grouping queries by step size improves efficiency.

What is the time complexity of XOR After Range Multiplication Queries II?

The optimal solution for XOR After Range Multiplication Queries II runs in O(n) time and uses O(n) space. Efficiently groups queries, reducing redundant updates and achieving linear complexity.

What is the brute force solution for XOR After Range Multiplication Queries II?

The brute force approach for XOR After Range Multiplication Queries II is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach directly applies each query to the array, iterating through the specified range and updating elements. It's straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve XOR After Range Multiplication Queries II?

XOR After Range Multiplication Queries II uses the following concepts: Array, Divide and Conquer. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for XOR After Range Multiplication Queries II?

Clarify the problem constraints before starting. Discuss potential optimizations during the interview. Practice similar problems to build confidence.

What are common mistakes when solving XOR After Range Multiplication Queries II?

Not handling modulo operations correctly. Forgetting to reset indices for each query.

#3655

XOR After Range Multiplication Queries II

Hard

ArrayDivide and ConquerHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This method groups queries based on the step size k, allowing efficient updates using a difference array approach. It minimizes redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: For each query, group by (k, l % k) and maintain a diff array for multipliers.
2Step 2: For each group, apply the multipliers to the corresponding indices in nums.
3Step 3: Compute the final XOR of the updated nums.

solution.py11 lines

1def xor_after_queries(nums, queries):
2    from collections import defaultdict
3    B = int(len(nums)**0.5)
4    groups = defaultdict(list)
5    for l, r, k, v in queries:
6        groups[(k, l % k)].append((l, r, v))
7    for (k, mod), group in groups.items():
8        for l, r, v in group:
9            for idx in range(l, r + 1, k):
10                nums[idx] = (nums[idx] * v) % (10**9 + 7)
11    return reduce(lambda x, y: x ^ y, nums)

ℹ

Complexity note: Efficiently groups queries, reducing redundant updates and achieving linear complexity.

1Grouping queries by step size improves efficiency.
2Using a difference array allows batch updates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.