<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:
        s = list(s)
        n = len(s)
        
        # Initialize pointers
        start = max(s)
        i = s.index(start)
        j = i + 1
        
        while j < n:
            # Find the next candidate where s[j] >= s[i]
            while j < n and s[j] < s[i]:
                j += 1
            
            if j >= n:
                # If we have reached the end, return the substring from i
                return "".join(s[i:])
            
            # Compare substrings starting from i and j
            l, r = i + 1, j + 1
            while r < n and s[l] == s[r] and l < j:
                l += 1
                r += 1
            
            if r == n:
                # If r reached the end, the substring from i is the result
                return "".join(s[i:])
            
            # Update the index i or j based on the comparison
            if s[l] < s[r]:
                i = j
            else:
                j = r
        
        return "".join(s[i:])
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