<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:
        # traverse string backwards
        # to max lexi, see if prev character is greater than current char - if so, then add to substring
        # keep track of biggest char in substring (this should start our substring)
        # also keep track of index of start of substring so that we know what to return
        # update these as we traverse to have a final substring to return
        # time: O(n)
        # space: O(1)
        biggest = s[-1]
        index = len(s) - 1
        for i in range(len(s) - 2, -1, -1):
            if s[i] > biggest: # if we have a new biggest char, then update it and make sure our substring starts here
                biggest = s[i]
                index = i
            elif s[i] == biggest: # if we encounter the biggest again, then check if substring starting at i > substring starting at index - if so then only update
                if s[i:len(s)] > s[index:len(s)]:
                    biggest = s[i]
                    index = i
        return s[index:len(s)]

            
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