<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:
        start = 0
        if len(s) <= 1:
            return s
        # fast = 1
        # new_start = None
        # new_fast = None
        # while True:
        #     if s[fast] > s[start]:
        #         start = fast
        #     elif s[fast] == s[start]:
        #         if new_start is None:
        #             new_start = fast
        #         else:
        d = []
        prev = None
        for i, j in enumerate(s):
            if prev is None:
                prev = j
                d.append(i)
            elif prev < j:
                prev = j
                d =[i]
            elif prev == j:
                d.append(i)
        left = None
        right = None
        if len(d) == 1:
            return s[d[0]:]
        res = s[d[0]:]
        for i in d[1:]:
            if s[i:] > res:
                res = s[i:]
        return res        






        #     fast += 1
        #     if fast >= len(s):
        #         break
        # return s[start:]        

        
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