<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:
        length = len(s)
        max_letter = max(s)
       
        

        start = s.index(max_letter)

        i = start+1

        while i < length:
            while i < length and s[i] != max_letter:
                i += 1
            
            if i >= length:
                return s[start:]
            
            # print(f"{i=}")
            
            left = start + 1
            right = i + 1
            while left < i and right < length and s[left] == s[right]:
                left += 1
                right += 1

            if right == length:
                return s[start:]
            
            if s[right] > s[left]:
                start = i
            else:
                i = right

                
        
        return s[start:]

                





        


       

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