How do you solve Maximum XOR Score Subarray Queries on LeetCode?

Maximum XOR Score Subarray Queries (LeetCode #3277) 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: Understanding XOR properties can lead to efficient solutions.

What is the time complexity of Maximum XOR Score Subarray Queries?

The optimal solution for Maximum XOR Score Subarray Queries runs in O(n) time and uses O(n) space. The time complexity is O(n) for preprocessing the prefix XOR array, and each query can be answered in O(1) time, making it efficient overall.

What is the brute force solution for Maximum XOR Score Subarray Queries?

The brute force approach for Maximum XOR Score Subarray Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray within the given range and calculating their XOR scores. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum XOR Score Subarray Queries?

Maximum XOR Score Subarray Queries uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Prefix Sum, Dynamic Programming.

What are the interview tips for Maximum XOR Score Subarray Queries?

Explain your thought process clearly. Consider edge cases, like small arrays or all zeros. Discuss time and space complexity before coding.

What are common mistakes when solving Maximum XOR Score Subarray Queries?

Not considering all subarrays in brute force. Overlooking the properties of XOR when optimizing.

#3277

Maximum XOR Score Subarray Queries

Hard

ArrayDynamic ProgrammingPrefix SumDynamic Programming

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)

By precomputing the XOR for all subarrays, we can efficiently answer each query in constant time. This takes advantage of the properties of XOR and cumulative results.

⚙️

Algorithm

3 steps

1Step 1: Create an array 'prefixXor' where prefixXor[i] is the XOR of nums[0] to nums[i].
2Step 2: For each query, calculate the XOR of the subarray using the prefixXor array.
3Step 3: Track the maximum XOR score for all subarrays within the specified range.

solution.py17 lines

1def maxXorSubarray(nums, queries):
2    n = len(nums)
3    prefixXor = [0] * n
4    prefixXor[0] = nums[0]
5    for i in range(1, n):
6        prefixXor[i] = prefixXor[i - 1] ^ nums[i]
7    results = []
8    for l, r in queries:
9        max_xor = 0
10        for i in range(l, r + 1):
11            for j in range(i, r + 1):
12                current_xor = prefixXor[j]
13                if i > 0:
14                    current_xor ^= prefixXor[i - 1]
15                max_xor = max(max_xor, current_xor)
16        results.append(max_xor)
17    return results

ℹ

Complexity note: The time complexity is O(n) for preprocessing the prefix XOR array, and each query can be answered in O(1) time, making it efficient overall.

1Understanding XOR properties can lead to efficient solutions.
2Precomputing results can save time during query processing.

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