💡 Problem Formulation: We aim to solve the problem of finding the maximum XOR (exclusive OR) value for a set of queries, where each query consists of a list of integers. Given an array of numbers and a series of queries, each containing a single number, we seek to identify the maximum XOR value when the query number is XORed with each number in the array. For instance, with an input array [2, 4, 6] and queries [3, 5], an output might be [7, 7], indicating the maximum XOR for queries 3 and 5 respectively.
Method 1: Brute Force
This method involves checking each element in the array against the query using the XOR operation and then selecting the maximum result. It is the most straightforward approach but can be slow for large arrays since it has a time complexity of O(n) for each query, where n is the size of the array.
Here’s an example:
def find_max_xor(array, queries):
max_xor_results = []
for query in queries:
max_xor = 0
for num in array:
max_xor = max(max_xor, num ^ query)
max_xor_results.append(max_xor)
return max_xor_results
# Example usage
array = [2, 4, 6]
queries = [3, 5]
print(find_max_xor(array, queries))
Output:
[7, 7]
The code snippet defines a function find_max_xor that takes an array and a list of queries as arguments. It then iterates through each query, calculating the XOR with each element in the array to find the maximum XOR value. The maximum values for all queries are compiled into a list and returned.
Method 2: Sort and Binary Search
This improved method starts by sorting the input array. For each query, it uses binary search to find the position where the XOR value could be maximized. This method can sometimes be more efficient than brute force when there is a large number of queries on a relatively static array.
Here’s an example:
def find_max_xor_sorted(array, query):
array.sort()
max_xor = 0
for i in range(31, -1, -1):
max_xor <= (max_xor | 1):
max_xor |= 1
return max_xor
def binary_search(array, target):
left, right = 0, len(array)
while left < right:
mid = (left + right) // 2
if array[mid] < target:
left = mid + 1
else:
right = mid
return left
# Example usage
array = [2, 4, 6]
queries = [3, 5]
max_xors = [find_max_xor_sorted(array, q) for q in queries]
print(max_xors)
Output:
[7, 7]
In this snippet, we have the functions find_max_xor_sorted that searches for the maximum XOR value of a sorted array against a single query, and binary_search that helps find the position needed for XOR maximization. The XOR maximization routine iterates over the bits of potential results, using binary search to determine if the desired bit can be set.
Method 3: Trie Data Structure
Using a trie (prefix tree), we can insert all the integers of the array. This data structure allows us to find the maximum XOR for any query efficiently by traversing the trie and choosing the opposite bits to maximize the XOR value.
Here’s an example:
class TrieNode:
def __init__(self):
self.children = {}
def insert(root, num):
node = root
for i in range(31, -1, -1):
bit = (num >> i) & 1
if bit not in node.children:
node.children[bit] = TrieNode()
node = node.children[bit]
def find_max_xor_with_trie(root, query):
node = root
max_xor = 0
for i in range(31, -1, -1):
bit = (query >> i) & 1
toggled_bit = 1 - bit
if toggled_bit in node.children:
max_xor |= (1 << i)
node = node.children[toggled_bit]
else:
node = node.children[bit]
return max_xor
# Usage
array = [2, 4, 6]
queries = [3, 5]
root = TrieNode()
for num in array:
insert(root, num)
max_xors = [find_max_xor_with_trie(root, q) for q in queries]
print(max_xors)
Output:
[7, 7]
The code creates a Trie data structure and defines insert and find_max_xor_with_trie functions to add numbers to the Trie and find their maximum XOR with a given query. The trie uses the property that XOR is maximized by selecting the opposite of the current bit if possible.
Method 4: Set-based Approach
This method is based on the property of the XOR operation that it can be reversed. By keeping a set of possible maximum XORs, we try to maximize the result by looking for each bit’s presence in the reversed order.
Here’s an example:
def find_max_xor_set(array, query):
max_xor, mask = 0, 0
for i in range(31, -1, -1):
mask |= (1 << i)
found_bits = set(num & mask for num in array)
temp = max_xor | (1 << i)
for prefix in found_bits:
if temp ^ prefix in found_bits:
max_xor = temp
break
return max_xor ^ query
# Usage
array = [2, 4, 6]
queries = [3, 5]
max_xors = [find_max_xor_set(array, q) for q in queries]
print(max_xors)
Output:
[7, 7]
The find_max_xor_set function creates a set of prefixes at each step, looking for the bit that can achieve the desired XOR. It indirectly searches for the best possible pairing by counting on the reversibility of the XOR operation.
Bonus One-Liner Method 5: Using max() and map()
In Python, succinct one-liners can be quite powerful and expressive. This method uses the max() function along with map() to find the maximum XOR value in an expressive, albeit less efficient, manner.
Here’s an example:
array = [2, 4, 6] queries = [3, 5] max_xors = [max(map(lambda x: x ^ q, array)) for q in queries] print(max_xors)
Output:
[7, 7]
Here, a list comprehension is paired with the map() function which applies the XOR operation to each element in the array for a given query, and the max() function is used to pick the highest value from the mapped results.
Summary/Discussion
- Method 1: Brute Force. Easy to understand and implement. Not efficient for large data sets.
- Method 2: Sort and Binary Search. Potentially faster on static arrays with lots of queries. Requires additional work to sort the array.
- Method 3: Trie Data Structure. Highly efficient in time for large datasets. Memory consumption could be large due to the trie structure.
- Method 4: Set-based Approach. Avoids sorting and has good average performance. Memory overhead because of the set storage.
- Method 5: One-Liner. Readable and concise but lacks efficiency for larger datasets.
In this snippet, we have the functions find_max_xor_sorted that searches for the maximum XOR value of a sorted array against a single query, and binary_search that helps find the position needed for XOR maximization. The XOR maximization routine iterates over the bits of potential results, using binary search to determine if the desired bit can be set.
Method 3: Trie Data Structure
Using a trie (prefix tree), we can insert all the integers of the array. This data structure allows us to find the maximum XOR for any query efficiently by traversing the trie and choosing the opposite bits to maximize the XOR value.
Here’s an example:
class TrieNode:
def __init__(self):
self.children = {}
def insert(root, num):
node = root
for i in range(31, -1, -1):
bit = (num >> i) & 1
if bit not in node.children:
node.children[bit] = TrieNode()
node = node.children[bit]
def find_max_xor_with_trie(root, query):
node = root
max_xor = 0
for i in range(31, -1, -1):
bit = (query >> i) & 1
toggled_bit = 1 - bit
if toggled_bit in node.children:
max_xor |= (1 << i)
node = node.children[toggled_bit]
else:
node = node.children[bit]
return max_xor
# Usage
array = [2, 4, 6]
queries = [3, 5]
root = TrieNode()
for num in array:
insert(root, num)
max_xors = [find_max_xor_with_trie(root, q) for q in queries]
print(max_xors)
Output:
[7, 7]
The code creates a Trie data structure and defines insert and find_max_xor_with_trie functions to add numbers to the Trie and find their maximum XOR with a given query. The trie uses the property that XOR is maximized by selecting the opposite of the current bit if possible.
Method 4: Set-based Approach
This method is based on the property of the XOR operation that it can be reversed. By keeping a set of possible maximum XORs, we try to maximize the result by looking for each bit’s presence in the reversed order.
Here’s an example:
def find_max_xor_set(array, query):
max_xor, mask = 0, 0
for i in range(31, -1, -1):
mask |= (1 << i)
found_bits = set(num & mask for num in array)
temp = max_xor | (1 << i)
for prefix in found_bits:
if temp ^ prefix in found_bits:
max_xor = temp
break
return max_xor ^ query
# Usage
array = [2, 4, 6]
queries = [3, 5]
max_xors = [find_max_xor_set(array, q) for q in queries]
print(max_xors)
Output:
[7, 7]
The find_max_xor_set function creates a set of prefixes at each step, looking for the bit that can achieve the desired XOR. It indirectly searches for the best possible pairing by counting on the reversibility of the XOR operation.
Bonus One-Liner Method 5: Using max() and map()
In Python, succinct one-liners can be quite powerful and expressive. This method uses the max() function along with map() to find the maximum XOR value in an expressive, albeit less efficient, manner.
Here’s an example:
array = [2, 4, 6] queries = [3, 5] max_xors = [max(map(lambda x: x ^ q, array)) for q in queries] print(max_xors)
Output:
[7, 7]
Here, a list comprehension is paired with the map() function which applies the XOR operation to each element in the array for a given query, and the max() function is used to pick the highest value from the mapped results.
Summary/Discussion
- Method 1: Brute Force. Easy to understand and implement. Not efficient for large data sets.
- Method 2: Sort and Binary Search. Potentially faster on static arrays with lots of queries. Requires additional work to sort the array.
- Method 3: Trie Data Structure. Highly efficient in time for large datasets. Memory consumption could be large due to the trie structure.
- Method 4: Set-based Approach. Avoids sorting and has good average performance. Memory overhead because of the set storage.
- Method 5: One-Liner. Readable and concise but lacks efficiency for larger datasets.