<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)
        first_zero_idx = -1
        number_of_zeros = 0
        for (i, b) in enumerate(binary):
            if b == "0":
                if first_zero_idx < 0:
                    first_zero_idx = i
                number_of_zeros += 1

        # print(number_of_zeros)
        if number_of_zeros <= 1:
            # There is no way we can improve in this case
            # since we only have one 0 in the binary string
            return binary

        transformed_zero_index = first_zero_idx + number_of_zeros - 1
        res = ["1" for _ in range(n)]
        res[transformed_zero_index] = "0"
        return "".join(res)
18900