<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 Solution {
private:
    void bitFunc(int target, vector<int>&nums, int &ans) {
        unordered_set<int>isPresent ;

        for(auto digit : nums) 
        {
            int num = (target & digit) ;
            if(isPresent.find(num ^ target) != isPresent.end()) 
            {
                ans = target ;
                return ;
            }
            isPresent.insert(num) ;
        }
    }

public:
    int findMaximumXOR(vector<int>& nums) {
        int ans = 0 ;

        for(int bit=31 ; bit>=0 ; bit--) 
        {
            bitFunc((ans | (1<<bit)), nums, ans) ;
        }

        return ans ;
    }
};
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