<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:
        n = len(s)
        curMax = max(s)

        candidates = [i for i, c in enumerate(s) if c == curMax]
        offset = 1
        
        while len(candidates) > 1:
            curMax = max([s[i+offset] for i in candidates if i+offset < n])
            new_candidates = []
            for i, start in enumerate(candidates):
                if i > 0 and candidates[i-1] + offset == start:
                    continue
                
                if start + offset < n and s[start + offset] == curMax:
                    new_candidates.append(start)
            candidates = new_candidates
            offset += 1
        
        return s[candidates[0]:]
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