<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:
    // for checking and setting jth bit in number n
    bool isset(int n, int j){return n&(1<<j);}
    void setbit(int& n, int j){n|=(1<<j);}

public:
    int findMaximumXOR(vector<int>& nums) {
        
        int n = nums.size();
        int xorTillNow = 0;
        vector<int> mask(n,0);

        for(int i=31;i>=0;i--)
        {
            // check if the ith bit can be set.
            //
            unordered_map<int,bool> ma;
            bool canSet = false;

            for(int j=0;j<n;j++)
            {
                mask[j] = mask[j]<<1;
                if(isset(nums[j],i))
                mask[j] = mask[j] | 1;
                ma[mask[j]]=1;
            }

            xorTillNow = (xorTillNow << 1);

            for(int j=0;j<n;j++)
            {
                int req = mask[j] xor (xorTillNow | 1);
                if(ma.count(req)) canSet = true;
                if(canSet) break;
            }

            

            if(canSet)
            {
                xorTillNow = (xorTillNow) | 1;
            }
        }

        return xorTillNow;
    }
};
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