<p>You are given an array of <code>logs</code>. Each log is a space-delimited string of words, where the first word is the <strong>identifier</strong>.</p>

<p>There are two types of logs:</p>

<ul>
	<li><b>Letter-logs</b>: All words (except the identifier) consist of lowercase English letters.</li>
	<li><strong>Digit-logs</strong>: All words (except the identifier) consist of digits.</li>
</ul>

<p>Reorder these logs so that:</p>

<ol>
	<li>The <strong>letter-logs</strong> come before all <strong>digit-logs</strong>.</li>
	<li>The <strong>letter-logs</strong> are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.</li>
	<li>The <strong>digit-logs</strong> maintain their relative ordering.</li>
</ol>

<p>Return <em>the final order of the logs</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> logs = [&quot;dig1 8 1 5 1&quot;,&quot;let1 art can&quot;,&quot;dig2 3 6&quot;,&quot;let2 own kit dig&quot;,&quot;let3 art zero&quot;]
<strong>Output:</strong> [&quot;let1 art can&quot;,&quot;let3 art zero&quot;,&quot;let2 own kit dig&quot;,&quot;dig1 8 1 5 1&quot;,&quot;dig2 3 6&quot;]
<strong>Explanation:</strong>
The letter-log contents are all different, so their ordering is &quot;art can&quot;, &quot;art zero&quot;, &quot;own kit dig&quot;.
The digit-logs have a relative order of &quot;dig1 8 1 5 1&quot;, &quot;dig2 3 6&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> logs = [&quot;a1 9 2 3 1&quot;,&quot;g1 act car&quot;,&quot;zo4 4 7&quot;,&quot;ab1 off key dog&quot;,&quot;a8 act zoo&quot;]
<strong>Output:</strong> [&quot;g1 act car&quot;,&quot;a8 act zoo&quot;,&quot;ab1 off key dog&quot;,&quot;a1 9 2 3 1&quot;,&quot;zo4 4 7&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= logs.length &lt;= 100</code></li>
	<li><code>3 &lt;= logs[i].length &lt;= 100</code></li>
	<li>All the tokens of <code>logs[i]</code> are separated by a <strong>single</strong> space.</li>
	<li><code>logs[i]</code> is guaranteed to have an identifier and at least one word after the identifier.</li>
</ul>

class Log{
    public:
    string content;
    int start;
    bool is_letter;
    Log(string s, set<char> &alphabet){
        content = s;
        start = -1;
        is_letter = false;
        int i(0);
        while(i<content.length()){
            if(content[i]==' ' && start==-1) start=i;
            if (start>-1 && alphabet.find(content[i])!=alphabet.end()) is_letter = true;
            ++i;           
        }
    }
    bool is_letter_log(){ return is_letter;}
};


class Solution {
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
        set<char> alphabet;
        for (char c('a'); c<='z'; ++c) alphabet.insert(c);
        vector<Log>  letter_log;
        vector<string> digit_log;
        for (int idx(0); idx<logs.size(); ++idx){
            Log tmp(logs[idx], alphabet);
            if(tmp.is_letter_log()) letter_log.push_back(tmp);
            else digit_log.push_back(logs[idx]);
        }
        auto comp = [&](Log &a, Log &b){
            auto & ac = a.content;
            int as = a.start;
            auto & bc = b.content;
            int bs = b.start;
            if (ac.substr(as, ac.length()-as+1) == bc.substr(bs, bc.length()-bs+1)) return ac.substr(0, as) < bc.substr(0, bs);
            return (ac.substr(as, ac.length()-as+1) < bc.substr(bs, bc.length()-bs+1));     
        };
        sort(letter_log.begin(), letter_log.end(), comp);
        vector<string> vec2;
        for (int k(0); k<letter_log.size(); ++k) vec2.push_back(letter_log[k].content);
        vec2.insert(vec2.end(), digit_log.begin(), digit_log.end());
        
        return vec2;
        
        
    }
};
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