<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:
        binary = []
        while n:
            binary.append(n % 2)
            n = n // 2
        k = len(binary)

        count = [0] * (k + 1)
        count[0], count[1] = 1, 2
        for i in range(2, k + 1):
            count[i] = count[i - 1] + count[i - 2]
        
        binary.reverse()
        print(binary)
        ans = 0
        for i, c in enumerate(binary):
            if c == 1:
                ans += count[k - i - 1]
                if i > 0 and binary[i - 1] == 1:
                    return ans
        
        return ans + 1
16600