421. Maximum XOR of Two Numbers in an Array

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

Constraints:

• 1 <= nums.length <= 2 * 105

• 0 <= nums[i] <= 231 - 1

Intuition

Approach:

Insert all the elements in Trie,
Then, for each element again , 
check if opposite bits exist and move there 

This way you generate answer 


9 -> 1001
Insert from start-> Start bits are valuable for maximum answer

Links

https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/description/

Video Links

https://www.youtube.com/watch?v=EIhAwfHubE8&list=PLgUwDviBIf0pcIDCZnxhv0LkHf5KzG9zp&index=6&ab_channel=takeUforward

Approach 1:

C++

struct Node{
    Node* links[2];
};

class Trie{
public:
    Node * root = new Node();

    void insert(int num){
        Node *temp = root;
        for(int i=31; i>=0; i--){
            int bit = (num>>i) & 1;
            if(temp->links[bit] == nullptr)
                temp->links[bit] = new Node();
            
            temp = temp->links[bit];
        }
    }
    
    int check(int num){
        Node* temp = root;
        int ans = 0;
        for(int i=31; i>=0; i--){
            int bit = (num>>i) & 1;
            if(temp->links[1-bit] == nullptr){
                temp = temp->links[bit];
            }
            else{
                temp = temp->links[1-bit];
                ans = ans | 1<<i;
            }
        }

        return ans;
    }
};

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        Trie trie;
        int ans = 0;
        for(auto &it: nums)
            trie.insert(it);

        for(auto &it: nums)
            ans = max(ans, trie.check(it));

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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