<p>You are given a binary string <code>binary</code> consisting of only <code>0</code>&#39;s or <code>1</code>&#39;s. You can apply each of the following operations any number of times:</p>

<ul>
	<li>Operation 1: If the number contains the substring <code>&quot;00&quot;</code>, you can replace it with <code>&quot;10&quot;</code>.

	<ul>
		<li>For example, <code>&quot;<u>00</u>010&quot; -&gt; &quot;<u>10</u>010</code>&quot;</li>
	</ul>
	</li>
	<li>Operation 2: If the number contains the substring <code>&quot;10&quot;</code>, you can replace it with <code>&quot;01&quot;</code>.
	<ul>
		<li>For example, <code>&quot;000<u>10</u>&quot; -&gt; &quot;000<u>01</u>&quot;</code></li>
	</ul>
	</li>
</ul>

<p><em>Return the <strong>maximum binary string</strong> you can obtain after any number of operations. Binary string <code>x</code> is greater than binary string <code>y</code> if <code>x</code>&#39;s decimal representation is greater than <code>y</code>&#39;s decimal representation.</em></p>

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

<pre>
<strong>Input:</strong> binary = &quot;000110&quot;
<strong>Output:</strong> &quot;111011&quot;
<strong>Explanation:</strong> A valid transformation sequence can be:
&quot;0001<u>10</u>&quot; -&gt; &quot;0001<u>01</u>&quot; 
&quot;<u>00</u>0101&quot; -&gt; &quot;<u>10</u>0101&quot; 
&quot;1<u>00</u>101&quot; -&gt; &quot;1<u>10</u>101&quot; 
&quot;110<u>10</u>1&quot; -&gt; &quot;110<u>01</u>1&quot; 
&quot;11<u>00</u>11&quot; -&gt; &quot;11<u>10</u>11&quot;
</pre>

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

<pre>
<strong>Input:</strong> binary = &quot;01&quot;
<strong>Output:</strong> &quot;01&quot;
<strong>Explanation:</strong>&nbsp;&quot;01&quot; cannot be transformed any further.
</pre>

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

<ul>
	<li><code>1 &lt;= binary.length &lt;= 10<sup>5</sup></code></li>
	<li><code>binary</code> consist of <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>
</ul>

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        n = len(binary)
        
        # Count the number of zeros
        zeros = binary.count('0')
        
        # If there are less than 2 zeros, we can't apply any operations
        if zeros < 2:
            return binary
        
        # Find the position of the first '0'
        first_zero = binary.index('0')
        
        # Calculate the position of the single '0' in the final string
        # It will be at (first_zero + zeros - 1)
        zero_pos = first_zero + zeros - 1
        
        # Construct the maximum binary string
        result = ['1'] * n
        result[zero_pos] = '0'
        
        return ''.join(result)


        #temp_binary_0 = temp_binary_1 = binary
        #l = len(binary)
        #for i in range(l):
        #    print(binary[i:i+2])
        #    if binary[i:i+2] == '00':
        #        temp_binary_0 = temp_binary_0[:i] + '10' + temp_binary_0[i+2:] 
        #    if binary[i:i+2] == '10':
        #        temp_binary_1 = temp_binary_1[:i] + '01' + temp_binary_1[i+2:] 
        #    if int(temp_binary_0) > int(binary):
        #        binary = temp_binary_0
        #    if int(temp_binary_1) > int(binary):
        #        binary = temp_binary_1
        #    print(binary)
#
        #    
        #return binary 
18200