<p>Given a string <code>s</code>, return <em>the last substring of</em> <code>s</code> <em>in lexicographical order</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abab&quot;
<strong>Output:</strong> &quot;bab&quot;
<strong>Explanation:</strong> The substrings are [&quot;a&quot;, &quot;ab&quot;, &quot;aba&quot;, &quot;abab&quot;, &quot;b&quot;, &quot;ba&quot;, &quot;bab&quot;]. The lexicographically maximum substring is &quot;bab&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;
<strong>Output:</strong> &quot;tcode&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 4 * 10<sup>5</sup></code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution:
    def lastSubstring(self, s: str) -> str:
        chars, max_char = list(s), ""
        max_char_indices = set()
        for idx, char in enumerate(chars):
            if char == max_char:
                max_char_indices.add(idx)

            if char > max_char:
                max_char = char
                max_char_indices.clear()
                max_char_indices.add(idx)

        last_substring = ""
        for start_idx in max_char_indices:
            current_substring = s[start_idx:]
            if current_substring > last_substring:
                last_substring = current_substring

        return last_substring

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