# Problem Understanding In this problem, we are given a list of $n$ positive integers representing the magic numbers of spells. The goal is to choose two distinct spells whose magic numbers, when combined with the bitwise XOR operation, yield the maximum possible result. The crucial aspects to consider include: - The XOR operation is bitwise, meaning it compares corresponding bits of two numbers. - We need to maximize the contribution from the most significant bits, as these bits weigh more in terms of the numerical value. - The input size can be as large as $10^5$, so an efficient algorithm is required. # Solution Analysis A brute-force approach would involve checking every pair of spells, but with $O(n^2)$ comparisons, this becomes impractical for large $n$. A more efficient technique takes advantage of properties of the XOR operation and uses a Trie (prefix tree): - **Trie (Bitwise Trie):** Build a Trie where each node represents a bit (from the most significant bit to the least significant bit) of the numbers. - **Insertion:** For each magic number, insert its binary representation into the Trie. - **Querying:** For each number, traverse the Trie to search for the number that gives the maximum possible XOR with it. When traversing for a particular bit, choose the opposite bit (if available) because XOR of different bits yields 1, thereby contributing to a higher overall value. - **Maximization:** Compute the XOR value for the current number with the best candidate found in the Trie and update the maximum if this value is greater than the current maximum. This approach leverages the bit-level structure of the numbers, ensuring we always try to maximize the bit at each level from the most significant to the least significant. # Time and Space Complexity Analysis - **Time Complexity:** - Inserting each number into the Trie takes $O(L)$ time, where $L$ is the number of bits in the numbers (typically $L \approx 32$ for 32-bit integers or $L \approx 64$ for 64-bit integers). For $n$ numbers, the total insertion becomes $O(n \cdot L)$. - Querying for the maximum XOR for each number also takes $O(L)$ per query, giving $O(n \cdot L)$ for all numbers. - Overall, the time complexity is $O(n \cdot L)$, which is efficient given that $L$ is a constant relative to $n$. - **Space Complexity:** - Each number contributes at most $L$ nodes to the Trie in the worst case. Therefore, the space complexity is $O(n \cdot L)$. - As $L$ is constant, the space requirement is linear in terms of $n$.