Solution Explanation

This problem asks to find the maximum XOR value between any two numbers in a given array. A brute-force approach would have O(n²) time complexity, which is inefficient for large arrays. The optimal solution leverages a Trie (prefix tree) data structure to efficiently search for numbers that maximize the XOR result.

Approach:

1. Trie Construction: A Trie is built to store the binary representation of each number in the input array. Each node in the Trie represents a bit (0 or 1), and the path from the root to a leaf node represents a number.

2. XOR Maximization: For each number x in the array, the Trie is traversed to find the number y that maximizes x XOR y. The traversal strategy is crucial: at each bit position, we try to choose the opposite bit (if it exists in the Trie) to maximize the XOR result. If the opposite bit doesn't exist, we take the same bit.

3. Maximum Value Tracking: The maximum XOR value found during the traversal is updated and returned as the final result.

Time Complexity Analysis:

• Trie Construction: Inserting each number into the Trie takes O(log₂(max(nums))) time, where max(nums) is the maximum number in the array. Since we have n numbers, the total time for construction is O(n * log₂(max(nums))).

• XOR Maximization: Searching for the optimal y for each x takes O(log₂(max(nums))) time. With n numbers, the total search time is O(n * log₂(max(nums))).

Therefore, the overall time complexity is O(n * log₂(max(nums))), which is significantly better than the brute-force O(n²). The space complexity is O(n * log₂(max(nums))) due to the Trie's size. In the worst-case scenario (all numbers are distinct and close to 231-1 ), the space complexity can be significant. However, the complexity is still linear in the number of input numbers, but logarithmic in the magnitude of input numbers.

Code Explanation (Python):

The Python code demonstrates the Trie implementation and the search algorithm.

class Trie:
    # ... (Trie node implementation - see full code above) ...
 
class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        trie = Trie()
        for x in nums:
            trie.insert(x)  # Build the Trie
        return max(trie.search(x) for x in nums) # Find max XOR for each number

The Trie class has methods for inserting numbers and searching for the number that maximizes XOR with a given number. The Solution class builds the Trie and then iterates through the array, finding the maximum XOR value for each number using the Trie. The max() function finds the overall maximum.

The other language implementations (Java, C++, Go, Rust) follow a similar structure, adapting the Trie implementation and syntax to their respective languages. They all share the same algorithmic approach and time/space complexity.