How do you solve Maximum Subarray XOR with Bounded Range on LeetCode?

Maximum Subarray XOR with Bounded Range (LeetCode #3845) 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: XOR operation properties can be leveraged for maximum values.

What is the time complexity of Maximum Subarray XOR with Bounded Range?

The optimal solution for Maximum Subarray XOR with Bounded Range runs in O(n) time and uses O(n) space. Linear complexity due to the single pass through the array and efficient Trie operations.

What is the brute force solution for Maximum Subarray XOR with Bounded Range?

The brute force approach for Maximum Subarray XOR with Bounded Range is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible subarrays and compute their XOR values. 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 Subarray XOR with Bounded Range?

Maximum Subarray XOR with Bounded Range uses the following concepts: Array, Bit Manipulation, Trie, Queue, Sliding Window, Prefix Sum, Monotonic Queue. The recommended patterns to study are: .

What are the interview tips for Maximum Subarray XOR with Bounded Range?

Explain your thought process clearly. Discuss edge cases and constraints.

What are common mistakes when solving Maximum Subarray XOR with Bounded Range?

Not maintaining the sliding window correctly. Forgetting to check the max-min condition before calculating XOR.

#3845

Maximum Subarray XOR with Bounded Range

Hard

ArrayBit ManipulationTrieQueueSliding WindowPrefix SumMonotonic Queue

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)

Use a sliding window to maintain valid subarrays and a Trie to efficiently compute maximum XOR values.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Trie and a sliding window with two pointers to track valid subarrays.
2Step 2: For each end index, add the prefix XOR to the Trie and check for maximum XOR with valid prefix XORs.
3Step 3: Adjust the start index to maintain the condition max - min <= k.

solution.py40 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.count = 0
5class Trie:
6    def __init__(self):
7        self.root = TrieNode()
8    def insert(self, num):
9        node = self.root
10        for i in range(15, -1, -1):
11            bit = (num >> i) & 1
12            if bit not in node.children:
13                node.children[bit] = TrieNode()
14            node = node.children[bit]
15            node.count += 1
16    def maxXOR(self, num):
17        node = self.root
18        max_xor = 0
19        for i in range(15, -1, -1):
20            bit = (num >> i) & 1
21            if 1 - bit in node.children:
22                max_xor |= (1 << i)
23                node = node.children[1 - bit]
24            else:
25                node = node.children[bit]
26        return max_xor
27
28def maxSubarrayXOR(nums, k):
29    trie = Trie()
30    max_xor = 0
31    prefix_xor = 0
32    start = 0
33    for end in range(len(nums)):
34        prefix_xor ^= nums[end]
35        trie.insert(prefix_xor)
36        while max(nums[start:end + 1]) - min(nums[start:end + 1]) > k:
37            prefix_xor ^= nums[start]
38            start += 1
39        max_xor = max(max_xor, trie.maxXOR(prefix_xor))
40    return max_xor

ℹ

Complexity note: Linear complexity due to the single pass through the array and efficient Trie operations.

1XOR operation properties can be leveraged for maximum values.
2Maintaining a sliding window helps in efficiently checking conditions.

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