<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:
        num = list(map(int, bin(n)[2:]))

        @cache
        def dp(pos: int, tight: bool, has_prev_one: bool) -> int:
            if pos == len(num): return 1
            res = 0
            bound = num[pos] if tight else 2
            for i in range(bound + 1):
                if i == 2: continue
                if i == 1:
                    if not has_prev_one: 
                        if i < bound: res += dp(pos + 1, False, True)
                        elif i == bound: res += dp(pos + 1, True, True)
                else:
                    if i < bound: res += dp(pos + 1, False, False)
                    elif i == bound: res += dp(pos + 1, True, False)
            return res

        return dp(0, True, False)
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