<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 dp[33][2][2];
    int rec(string&s, int pos, int prev, bool tight){
        if(pos>=s.length())     return 1;
        if(dp[pos][prev][tight]!=-1)    return dp[pos][prev][tight];
        int ans=0;
        int ub= tight ? s[pos]-'0' : 1;
        int count=0;
        for(int i=0; i<=ub; i++){
            if(i==0 or (i==1 and prev!=1)){
                ans+=rec(s, pos+1, i, tight and i==ub);
            }
        }
        return dp[pos][prev][tight]=ans;
    }
    int findIntegers(int n) {
        string s="";
        while(n){
            s=to_string(n&1)+s;
            n>>=1;
        }
        memset(dp, -1, sizeof(dp));
        return rec(s, 0, 0, true);
    }
};


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