<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)
        maxChar = max(s)

        candidates = [i for i, ch in enumerate(s) if ch == maxChar]
        offset = 1
        
        while len(candidates) > 1:
            newCandidates = []
            currMax = max([s[idx + offset] for idx in candidates if idx+offset < len(s)])
            for i, idx in enumerate(candidates):
                if i > 0 and candidates[i - 1] + offset == idx:
                    continue
                if idx + offset < n and s[idx + offset] == currMax:
                    newCandidates.append(idx)

            candidates = newCandidates
            offset += 1
        
        return s[candidates[0]:]
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