<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:
        c = ord('z')
        while c >= ord('a'):
            substrs = s.split(chr(c))
            if len(substrs) == 1:
                c -= 1
            else:
                # i = 0
                # while substrs[-1] == '' and len(substrs) > 2:
                #     substrs.pop(-1)
                #     i += 1
                # substrs[-1] += chr(c) * i

                # if substrs[-1] == '' and len(substrs) > 2:
                #     substrs.pop(-1)
                # substrs[-1] += chr(c)

                i = len(substrs) - 2
                while i >= 1:
                    if substrs[i] == '':
                        substrs.pop(i)
                        substrs[i] = chr(c) + substrs[i]
                    i -= 1

                substr_idx = substrs[1:].index(sorted(substrs[1:])[-1]) + 1
                split_idx = sum([len(substr) for substr in substrs][:substr_idx]) + substr_idx - 1

                if substrs[-1] == '' and len(substrs) > 2:
                    print(chr(c) + substrs[-2] + chr(c))
                    if chr(c) + substrs[-2] + chr(c) > s[split_idx:]:
                        return chr(c) + substrs[-2] + chr(c)
                return s[split_idx:]
798