<p>Given an integer array <code>nums</code>, return <em>the maximum result of </em><code>nums[i] XOR nums[j]</code>, where <code>0 &lt;= i &lt;= j &lt; n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [14,70,53,83,49,91,36,80,92,51,66,70]
<strong>Output:</strong> 127
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Node{
    public:
    Node* links[2];
    Node(){
        links[0]=NULL;
        links[1]=NULL;
    }
};

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        Node* root=new Node();
        for(auto x:nums){
            Node* curr=root;
            string s="";
            for(int i=31;i>=0;i--){
                if(x&(1<<i))s+='1';
                else s+='0';
            }
            for(int i=0;i<s.size();i++){
                if(curr->links[s[i]-'0']==NULL){
                    curr->links[s[i]-'0']=new Node();
                }
                curr=curr->links[s[i]-'0'];
            }
        }
        int ans=0;
        for(auto x:nums){
            Node* curr=root;
            string s="";
            for(int i=31;i>=0;i--){
                if(x&(1<<i))s+='1';
                else s+='0';
            }
            int val=0;
            for(int i=0;i<s.size();i++){
                if(s[i]=='1'){
                    if(curr->links[0]!=NULL){
                        curr=curr->links[0];
                        val|=(1<<(31-i));
                    }else{
                        curr=curr->links[1];
                    }
                }else{
                    if(curr->links[1]!=NULL){
                        curr=curr->links[1];
                        val|=(1<<(31-i));
                    }else{
                        curr=curr->links[0];
                    }
                }
            }
            ans=max(ans,val);
        }
        return ans;
    }
};
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