<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 num) {
    string bits;
    for (; num; num >>= 1)
      bits += to_string(num & 1);

    const int n = bits.length();
    vector<int> zero(n, 1);
    vector<int> one(n, 1);

    for (int i = 1; i < n; ++i) {
      zero[i] = zero[i - 1] + one[i - 1];
      one[i] = zero[i - 1];
    }

    int ans = zero[n - 1] + one[n - 1];

    for (int i = n - 2; i >= 0; --i) {
      // The numbers > num and <= 2^n - 1 are invalid.
      if (bits[i] == '1' && bits[i + 1] == '1')
        break;
      if (bits[i] == '0' && bits[i + 1] == '0')
        ans -= one[i];
    }

    return ans;
  }
};
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