<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:
            l, r =0, 1
            while r < len(s):
                print(l, r)
                if s[r] < s[l]:
                    r += 1
                    continue
                elif s[r] > s[l]:
                    l = r
                    r += 1
                    continue
                else:
                    rr = r + 1
                    broke = False
                    while rr - r <= r - l and rr <= len(s):
                        if s[l:l + (rr - r)] < s[r:rr]:
                            l = r
                            broke = True
                            break
                        elif s[l:l + (rr - r)] > s[r:rr]:
                            broke = True
                            break
                        else:
                            rr += 1
                    r += 1
                    if not broke:
                        r = rr - 1
            return s[l:]
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