<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 findIntegers(int n) {
        vector<int> fib(31, 0);  // Fibonacci sequence for binary length calculation
        fib[0] = 1;
        fib[1] = 2;
        for (int i = 2; i < 31; ++i) {
            fib[i] = fib[i - 1] + fib[i - 2];
        }
        
        int prev_bit = 0, result = 0;
        for (int i = 29; i >= 0; --i) {
            if (n & (1 << i)) {  // Check if the ith bit is set
                result += fib[i];
                if (prev_bit) return result;  // If there are two consecutive 1's, stop
                prev_bit = 1;
            } else {
                prev_bit = 0;
            }
        }
        return result + 1;  // Add 1 for the number itself
    }
};

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