<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:
    def findIntegers(self, n: int) -> int:
        f = [0] * 32
        f[0] = 1
        f[1] = 2
        for i in range(2, 32):
            f[i] = f[i-1] + f[i-2]
        
        res, prev_bit = 0, 0
        for i in range(30, -1, -1):
            if (n & (1 << i)) != 0: # 1
                res += f[i]
                if prev_bit == 1:
                    break
                prev_bit = 1
            else: # 0
                prev_bit = 0
            
            if i == 0:
                res += 1
        
        return res
16300