<p>Given an integer array <code>arr</code>, return <em>the number of distinct bitwise ORs of all the non-empty subarrays of</em> <code>arr</code>.</p>

<p>The bitwise OR of a subarray is the bitwise OR of each integer in the subarray. The bitwise OR of a subarray of one integer is that integer.</p>

<p>A <strong>subarray</strong> is a contiguous non-empty sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> arr = [0]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is only one possible result: 0.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,1,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
These yield the results 1, 1, 2, 1, 3, 3.
There are 3 unique values, so the answer is 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The possible results are 1, 2, 3, 4, 6, and 7.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:

    int subarrayBitwiseORs(vector<int>& arr){
        vector<vector<int>> pos(32);
        int n=arr.size();
        int cnt=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<32;j++){
                if((1<<j)&arr[i]){
                    pos[j].push_back(i);
                }
            }
        }
        unordered_set<int> nums;
        for(int i=0;i<n;i++){
            map<int,int> z;
            for(int j=0;j<32;j++){
                if(((1<<j)&arr[i])==0 && pos[j].size()){                    
                    int p=lower_bound(pos[j].begin(),pos[j].end(),i)-pos[j].begin();
                    if(p!=pos[j].size()) z[pos[j][p]]+=(1<<j);
                }
            }
            nums.insert(arr[i]);
            int y=arr[i];
            for(auto g:z){
                y+=g.second;
                nums.insert(y);
            }
        }
        return nums.size();
    }
};
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