<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:
    vector<vector<vector<int>>> memo;

    int dp(int pos, bool tight, bool prev_one, const string &binary) {
        if (pos == binary.size()) return 1;

        if (memo[pos][tight][prev_one] != -1) return memo[pos][tight][prev_one];

        int limit = tight ? binary[pos] - '0' : 1;
        int ans = 0;

        ans += dp(pos + 1, tight && (limit == 0), false, binary);

        if (!prev_one && limit >= 1) {
            ans += dp(pos + 1, tight && (limit == 1), true, binary);
        }

        return memo[pos][tight][prev_one] = ans;
    }

    int findIntegers(int n) {
        string binary = bitset<32>(n).to_string();
        int len = binary.size();

        int first_one = binary.find('1');
        binary = binary.substr(first_one);

        memo = vector<vector<vector<int>>>(len, vector<vector<int>>(2, vector<int>(2, -1)));

        return dp(0, true, false, binary);
    }
};

12000