<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) {
    const int K = 31;
    vector<vector<int>> index(K);
    const int n = arr.size();
    for (int i = 0; i < n; ++i) {
      for (int b = 0; b < K; ++b) {
        if ((arr[i] >> b) & 1) {
          index[b].push_back(i);
        }
      }
    }
    set<int> ans;
    for (int l = 0; l < n; ++l) {
      //cout << "l: " << l << endl;
      int cur = arr[l];
      ans.insert(cur);
      for (int r = l+1; r < n;) {
        //cout << "r: " << r << endl;
        int nx = n;
        for (int b = 0; b < K; ++b) {
          if ((cur >> b)&1) continue;
          auto it = lower_bound(index[b].begin(), index[b].end(), r);
          int temp = it == index[b].end() ? n : *it;
          nx = min(nx, temp);
        }
        if (nx >= n) {
          break;
        }
        cur |= arr[nx];
        ans.insert(cur);
        r = nx+1;
      }
    }
    return ans.size();
  }
};
137762