<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 {
public:
    string lastSubstring(string s) {
        int n=s.length();
    bool current=false;
    char x;
    int maxx=-1;
    int length=0;
    int best=-1;
    int first=0;
    for (int i=0;i<n;i++) {
        if (s[i]-'a'>maxx) {
            maxx=s[i]-'a';
            x=s[i];
            best=i;
        }
    }
    for (int i=0;i<n;i++) {
        if (s[i]==x) {
            first=i;
            break;
        }
    }
    for (int i=best+1;i<n;i++) {

        if (current) {
//            cout<<i<<' '<<s[i]<<el;
            if (s[best+length]-'a'<s[i]-'a') {
                best=i-length;
                current=false;
            }
            if (s[best+length]-'a'>s[i]-'a') {
                current=false;
            }
            if (s[best+length]-'a'==s[i]-'a') {
                length++;
            }
        }
        if (s[i]==x&&s[i-1]!=x) {
            current=true;
            length=1;

        }
    }
    string ans="";
    for (int i=best;i<n;i++) ans+=s[i];
    return ans;
    }
};
23100