<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 index , count;
        char po ;
        string st =""; string out = "" ;
        vector<int> isfound(26 , -1) ;
        for(int i =  s.size() -1 ; i >= 0 ; i--) isfound[s[i] -'a' ] = i ;
        for(char i = 25 ; i >=0 ; i--) if(isfound[i] != -1) {po = i ; break ;}
        for(int i = isfound[po] ; i < s.size() ; i++){
            out += s[i] ;
            st += s[i] ;
            if((s[i] - 'a' == po)){
                out = (out >= st) ? out : st ;
                st = s[i] ;
                while( (i < s.size()-1) && ( s[i] == s[i+1])){
                    out += s[i] ;
                    st += s[i] ;
                    i++ ;
                }
            }
        }
        out = (out >= st) ? out : st ;
        return out ;
    }
};
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