algoadvance

You are given an array of integers nums. A strong pair (i, j) is defined as follows:

Return the maximum XOR of any strong pair.

Example:

Input: nums = [3, 10, 5, 25, 2, 8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.

Clarifying Questions

  1. Can the input array contain negative numbers?
    • No, the problem as typically defined will involve non-negative integers.
  2. What are the constraints on the size of the array?
    • Usually, the size of the array will be such that a brute force approach might not be efficient. Let’s assume typical constraints like 1 <= nums.length <= 2 * 10^4 and 0 <= nums[i] <= 2^31 - 1.
  3. Should the solution handle special cases like an empty array?
    • Assumption is that the array contains at least two elements, given the requirement of finding a pair.

Strategy

To solve this problem, a brute force approach would involve checking all pairs (i, j) and computing their XOR value to find the maximum. However, with an upper limit of 2*10^4 on the length of nums, this approach would be inefficient since its time complexity would be (O(n^2)).

Instead, we can use a more efficient method leveraging bit manipulation and a trie data structure to store the binary representations of numbers. Here’s a step-by-step outline of the optimized approach:

  1. Convert numbers to binary and insert into a trie: We’ll create a binary trie where each bit of the number represents a path in the trie.

  2. Maximize XOR with greedy approach: For each number, we will try to find the maximum possible XOR by traversing the trie. The goal is to select paths in the trie that would maximize the bit difference at each level.

Code

Here’s an implementation of the solution using a trie:

class TrieNode:
    def __init__(self):
        self.children = dict()

class Trie:
    def __init__(self):
        self.root = TrieNode()
        
    def insert(self, num):
        node = self.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(self, num):
        node = self.root
        xor = 0
        for i in range(31, -1, -1):
            bit = (num >> i) & 1
            toggled_bit = 1 - bit
            if toggled_bit in node.children:
                xor |= (1 << i)
                node = node.children[toggled_bit]
            else:
                node = node.children.get(bit, node)
        return xor
        
class Solution:
    def findMaximumXOR(self, nums):
        trie = Trie()
        max_xor = 0
        for num in nums:
            trie.insert(num)
        for num in nums:
            max_xor = max(max_xor, trie.find_max_xor(num))
        return max_xor

Time Complexity

This approach ensures we handle large inputs efficiently and find the maximum XOR for any strong pair.

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