<p>Given the binary representation of an integer as a string <code>s</code>, return <em>the number of steps to reduce it to </em><code>1</code><em> under the following rules</em>:</p>

<ul>
	<li>
	<p>If the current number is even, you have to divide it by <code>2</code>.</p>
	</li>
	<li>
	<p>If the current number is odd, you have to add <code>1</code> to it.</p>
	</li>
</ul>

<p>It is guaranteed that you can always reach one for all test cases.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1101&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> &quot;1101&quot; corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14.&nbsp;
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.&nbsp; 
Step 5) 4 is even, divide by 2 and obtain 2.&nbsp;
Step 6) 2 is even, divide by 2 and obtain 1.&nbsp; 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;10&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> &quot;10&quot; corresponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.&nbsp; 
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1&quot;
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length&nbsp;&lt;= 500</code></li>
	<li><code>s</code> consists of characters &#39;0&#39; or &#39;1&#39;</li>
	<li><code>s[0] == &#39;1&#39;</code></li>
</ul>

class Solution {
public:
    int numSteps(string s) {
        int ans=0;
        reverse(s.begin(),s.end());
        while(s.size()!=1){
            ans++;
            int i=0;
            if(s[i]=='1'){
            while(s[i]!='0' && i<s.size()){
                s[i]='0';
                i++;
            }
            if(i==s.size()){
                s+='1';
            }
            else{
                s[i]='1';
            }
            }
            else{
                string temp=s.substr(1,s.length()-1);
                s=temp;
            }

        }
      
        return ans;
    }
};
11100