<p>Given a positive integer <code>n</code>, return the number of the integers in the range <code>[0, n]</code> whose binary representations <strong>do not</strong> contain consecutive ones.</p>

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

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong>
Here are the non-negative integers &lt;= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. 
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
int solve(int pos, bool prevOne, bool tight, string& binary, vector<vector<vector<int>>>& dp) {
   
    if (pos == binary.size()) {
        return 1;
    }
    if (dp[pos][prevOne][tight] != -1) {
        return dp[pos][prevOne][tight];
    }
    int limit = tight ? binary[pos] - '0' : 1; 
  int ans = 0;
      for (int digit = 0; digit <= limit; ++digit) {
        if (prevOne ==1&& digit == 1) {
            continue; 
        }   
        ans += solve(pos + 1, digit == 1, tight && (digit == limit), binary, dp);
      }
     return dp[pos][prevOne][tight] = ans;
    }


    int findIntegers(int n) {
        string binary = bitset<32>(n).to_string(); 
    binary = binary.substr(binary.find('1')); 
    int len = binary.size();
    vector<vector<vector<int>>> dp(len, vector<vector<int>>(2, vector<int>(2, -1)));

    return solve(0, false, true, binary, dp);
    }
};
11500