<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 Solution {
public:

class LetterLogComparator
{
public:
      bool operator()(const std::string& left, const std::string& right) {
        int l_space = left.find(' ');
        int r_space = right.find(' ');

        string l_contents = string(left.begin() + l_space + 1, left.end());
        string r_contents = string(right.begin() + r_space + 1, right.end());
        string l_ident = string(left.begin(), left.begin() + l_space);
        string r_ident = string(right.begin(), right.begin() + r_space);

        if (l_contents != r_contents) {
            return l_contents > r_contents;
        }
        else {
            return l_ident > r_ident;
        }
    }
};
    bool letterLogComparator(const std::string& left, const std::string& right) {
        int l_space = left.find(' ');
        int r_space = right.find(' ');

        string l_contents = string(left.begin() + l_space + 1, left.end());
        string r_contents = string(right.begin() + r_space + 1, right.end());
        string l_ident = string(left.begin(), left.begin() + l_space);
        string r_ident = string(right.begin(), right.begin() + r_space);

        if (l_contents != r_contents) {
            return l_contents > r_contents;
        }
        else {
            return l_ident > r_ident;
        }
    }
    
    bool isDigitLog(string s) {
        for (int i = 0; i < s.length(); ++i) {
            if (isalpha(s[i]))
                return false;
        }

        return true;
    }
    vector<string> reorderLogFiles(vector<string>& logs) {
        priority_queue<string, vector<string>, LetterLogComparator> letterPq;
        vector<string> digitLogs;
        // push letter logs
        for (int i = 0; i < logs.size(); ++i) {
            int space_idx = logs[i].find(' ');
            string contents = string(logs[i].begin() + space_idx + 1, logs[i].end());
            if (isDigitLog(contents)) {
                digitLogs.push_back(logs[i]);
            }  
            else {
                letterPq.push(logs[i]);
            }
        }


        vector<string> orderedLogs;
        while (!letterPq.empty()) {
            orderedLogs.push_back(letterPq.top());
            letterPq.pop();
        }

        for (int i = 0; i < digitLogs.size(); ++i) {
            orderedLogs.push_back(digitLogs[i]);
        }

        return orderedLogs;

    }
};
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