<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 helper(string s){
        // max stores the maximium character in the string 
        // maxIndex stores the starting index of the largest substring starting with the max character


        char max = s[0];
        int length = s.size(), maxIndex = 0;
        for(int i = 1 ; i < length; ++i){
            if ( s[i] > max ){
                maxIndex = i;
                max = s[i];
            }
            else if (s[i] == max){
                int j = 1;
                while (s[i + j] == s[maxIndex + j] && maxIndex + j < i ) j++;
                if (s[i + j] > s[maxIndex + j]) maxIndex = i;
                i = i + j-1;
            }
        }
        return s.substr(maxIndex,length - maxIndex + 1);

    }
    string lastSubstring(string s) {
        while(true){
            string res=helper(s);
            if(helper(res)==res) return res;
            else s=res;
        }
        return " ";
    }
};

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