<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:
    def numSteps(self, s: str) -> int:
        l = len(s)
        # 1101
        # 1 0 1 1
        last_one = s.rfind('1')
        if last_one == 0:
            return l - 1
        m = map(lambda x: 1 if x == '0' else 0, s[last_one::-1])
        # print(list(m))
        # print([n for n in m])
        # last_one = s.rfind('1')
        zeroes = sum(list(m))
        # print(zeroes)

        # zeroes = sum(map(lambda x: 1 if x == '0' else 0, s[last_one::-1]))
        # print(zeroes, l)
        return l + zeroes + 1# if zeroes else l - 1
        # very large numbers, so need to handle smartly
        steps = len(s) - 1
        carry = 0

        last_one_ix = s.rfind('1') #must be one
        next_zero_ix = s[:last_one_ix].rfind('0') #might not


        for b in s[:0:-1]:
            if b == '0' and carry:
                steps +=1
            elif b == '1' and not carry:
                carry = 1
                steps += 1

        if carry:
            steps += 1

        return steps 

       
16300