421 Maximum XOR of Two Numbers in an Array

421. Maximum XOR of Two Numbers in an Array

1. Question

Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai< 231.

Find the maximum result of aiXOR 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.

2. Implementation

(1) Brute Force

class Solution {
    public int findMaximumXOR(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int max = 0;

        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                max = Math.max(max, nums[i] ^ nums[j]);
            }
        }
        return max;
    }
}

(2) Bit Manipulation + Trie

class Solution {
    class TrieNode {
        TrieNode[] childNode;

        public TrieNode() {
            childNode = new TrieNode[2];
        }
    }

    class Trie {
        TrieNode root;

        public Trie() {
            root = new TrieNode();
        }

        public void insert(int num) {
            TrieNode curNode = root;
            for (int i = 31; i >= 0; i--) {
                int bit = (num >>> i) & 1;

                if (curNode.childNode[bit] == null) {
                    curNode.childNode[bit] = new TrieNode();
                }
                curNode = curNode.childNode[bit];
            }
        }
    }

    public int findMaximumXOR(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        Trie trie = new Trie();

        for (int num : nums) {
            trie.insert(num);
        }

        int max = 0;

        for (int num : nums) {
            int curSum = 0;
            TrieNode curNode = trie.root;

            for (int i = 31; i >= 0; i--) {
                int bit = (num >>> i) & 1;

                if (curNode.childNode[bit ^ 1] != null) {
                    curSum |= (1 << i);
                    curNode = curNode.childNode[bit ^ 1];
                }
                else {
                    curNode = curNode.childNode[bit];
                }
            }

            max = Math.max(max, curSum);
        }
        return max;
    }
}

3. Time & Space Complexity

Brute Force: 时间复杂度O(n^2), 空间复杂度O(1)

Bit Manipulation + Trie: 时间复杂度O(n), 空间复杂度O(n)