<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) {
        int n=arr.size();
        unordered_map<int,vector<int>>dp;
        for(int i=0;i<n;i++){
            for(int j=0;j<=31;j++){
                if((arr[i]>>j)&1){
                    dp[j].push_back(i);
                }
            }  
        }
        unordered_set<long long>unique_ors;
        for(int i=0;i<n;i++){
            map<int,long long>dp1;
            for(int j=0;j<=31;j++){
                int bit=(arr[i]>>j)&1;
                if(bit)continue;
                auto it=upper_bound(dp[j].begin(),dp[j].end(),i);
                if(it==dp[j].end())continue;
                dp1[*it]|=(1<<j);
            }
            unique_ors.insert((long long)arr[i]);
            long long y=0;
            for(auto ind:dp1){
                y|=ind.second;
                long long x=arr[i]|y;
                unique_ors.insert(x);
            }

        }
        return unique_ors.size();
    }
};

// 0100
// 0001
// 1111

// 10110101
//  1  1 1




// 111111111

// dp[ind][30].vector<>
257087