Last updated on July 18th, 2024 at 03:18 am

Here, We see Maximum XOR of Two Numbers in an Array LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Maximum XOR of Two Numbers in an Array LeetCode Solution

Problem Statement

Given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 <= i <= j < n.

Example 1:
Input: nums = [3,10,5,25,2,8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.

Example 2:
Input: nums = [14,70,53,83,49,91,36,80,92,51,66,70]
Output: 127

1. Maximum XOR of Two Numbers in an Array LeetCode Solution C++

class TrieNode{
public:
    TrieNode *child[2];
    TrieNode(){
        this->child[0] = NULL;
        this->child[1] = NULL;
    }
};

class Solution {
    TrieNode *newNode;
    void insert(int x){
        TrieNode *t = newNode;
        bitset<32> bs(x);
        for(int j=31; j>=0; j--){
            if(!t->child[bs[j]]) t->child[bs[j]] = new TrieNode();
            t = t->child[bs[j]];
        }    
    }
public:
    int findMaximumXOR(vector<int>& nums) {
        newNode = new TrieNode();
        for(auto &n : nums) insert(n);
        int ans=0;
        for(auto n : nums){
            ans = max(ans, maxXOR(n));
        }
        return ans;
    }
    int maxXOR(int n){
        TrieNode *t = newNode;
        bitset<32> bs(n);
        int ans=0; 
        for(int j=31; j>=0; j--){
            if(t->child[!bs[j]]) ans += (1<<j), t = t->child[!bs[j]];
            else t = t->child[bs[j]];
        }
        return ans;
    }
};

2. Maximum XOR of Two Numbers in an Array LeetCode Solution Java

class Node {
    HashMap<Integer, Node> children;
    Node() {
        this.children = new HashMap<>();
    }
}

class Trie {
    Node root;
    Trie() {
        this.root = new Node();
    }
    public void insert(int[] A) {
        for(int num : A) {
            Node curr = this.root;
            for(int i=31;i>=0;i--) {
                int currBit = (num >> i) & 1;
                if(!curr.children.containsKey(currBit)) 
                    curr.children.put(currBit, new Node());
                curr = curr.children.get(currBit);
            }
        }
    }
}

class Solution {
    public int findMaximumXOR(int[] nums) {
        Trie trie = new Trie();
        trie.insert(nums);
        int max = 0;
        for(int num : nums) {
            Node curr = trie.root;
            int currSum = 0;
            for(int i=31;i>=0;i--) {
                int requiredBit = 1-((num >> i) & 1);
                if(curr.children.containsKey(requiredBit)) {
                    currSum |= (1<<i);
                    curr = curr.children.get(requiredBit);
                } else {
                    curr = curr.children.get(1-requiredBit);
                }
            }
            max = Math.max(max, currSum);
        }
        return max;
    }
}

3. Maximum XOR of Two Numbers in an Array Solution JavaScript

var findMaximumXOR = function(nums) {
  let trie = new Trie();
  let root = trie.root;
  for(let num of nums) {
    trie.insert(num);
  }
  let ans = 0;
  let INTEGER_MAX = Math.pow(2, 31) - 1;
  for(let num of nums) {
    let target = num ^ INTEGER_MAX;
    let found = trie.search(target);
    ans = Math.max(ans, num ^ found);
  }
  return ans;
};

class Trie {
  root = new Map();
  insert(number) {
    let index = 30;
    let node = this.root;
    while(index >= 0) {
      let mask = 1 << index;
      let bit = (number & mask) > 0 ? 1 : 0;
      if(!node.has(bit)) {
        node.set(bit, new Map());
      }
      node = node.get(bit);
      index--;
    }
  }
  search(number) {
    let ans = 0;
    let index = 30;
    let node = this.root;
    while(index >= 0) {
      let mask = 1 << index;
      let bit = (number & mask) > 0 ? 1 : 0;
      if (node.has(bit)) {
        node = node.get(bit);
      } else {
        bit = bit === 0 ? 1 : 0
        node = node.get(bit);
      }
      if (bit === 1) {
        ans = ans + mask;
      }
      index--;
    }
    return ans;
  }
}

4. Maximum XOR of Two Numbers in an Array Solution Python

class Solution(object):
    def findMaximumXOR(self, nums):
        m, mask = 0, 0
        for i in range(32)[::-1]:
            mask |= 1 << i
            prefixes = {n & mask for n in nums}
            tmp = m | (1 << i)
            if any(prefix ^ tmp in prefixes for prefix in prefixes):
                m = tmp
        return m