## Maximum Xor of Two Numbers in an Array

Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai< 231. Find the maximum result of ai XOR aj, where 0 ≤ i, j < n. Could you do this in O(n) runtime? Example Input: [3, 10, 5, 25, 2, 8] Output: 28 Explanation: The maximum result is 5 ^ 25 = 28. Solution (prefix hashset - less efficient) Time: O(N) Space: O(1) class Solution: def findMaximumXOR(self, nums: List[int]) -> int: # 0bxxxxxx - 0b L = len(bin(max(nums))) - 2 max_xor = 0 for i in range(L-1, -1, -1): max_xor <<= 1 print("max_xor:", bin(max_xor) ) # set the rightmost bit to 1 curr_xor = max_xor | 1 print("curr_xor:", bin(curr_xor) ) # highest (L-i) bits prefixes = {num >> i for num in nums} print("prefixes:", [bin(p) for p in prefixes] ) # as long as there exists a p that makes curr_xor^p > 0 # Update max_xor, if two of these prefixes could result in curr_xor.