Non-negative Integers without Consecutive Ones
<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>
Dynamic Programming
class Solution:
    def findIntegers(self, n: int) -> int:
        @cache
        def dfs(i: int, pre: int, limit: bool) -> int:
            if i < 0:
                return 1
            up = (n >> i & 1) if limit else 1
            ans = 0
            for j in range(up + 1):
                if pre and j:
                    continue
                ans += dfs(i - 1, j, limit and j == up)
            return ans

        return dfs(n.bit_length() - 1, 0, True)
