<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 str) {
    int maxSubstringStart = 0;
    for (int i = 0; i < str.length();) {
        int j = i, k = i + 1;
        while (str[j] >= str[k] && k < str.length()) {
            if (str[j] > str[k]) {
                j = i;
                k++;
            } else {
                k++;
                j++;
            }
        }
        if (k == str.length()) {
            maxSubstringStart = i;
            break;
        } else {
            int diff = k -j;
            while (i+diff<=k){
                i+=diff;
            }
            maxSubstringStart = k;
        }
    }

    return str.substr(maxSubstringStart, str.length());
}


};
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