<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:
    vector<string> reorderLogFiles(vector<string>& logs) {
        int n  = logs.size();
        vector<string>let;
        vector<string>dig;
        
        for(int i = 0; i < n ; i++){
            int j = 0;
            while(logs[i][j]!=' '){
                j++;
            }
            if(logs[i][j+1] <='9' && logs[i][j+1]>='0'){
                dig.push_back(logs[i]);
            }
            else{
                let.push_back(logs[i]);
            }
        }
        
        for(int i = 0; i < let.size()-1; i++){
            for(int j = i+1; j < let.size() ; j++){
                int j1 = 0;
                while(let[i][j1] != ' '){
                    j1++;
                }
                string temp1 = let[i].substr(j1+1, let[i].length()-j1-1);
            
                int j2 = 0;
                while(let[j][j2] != ' '){
                    j2++;
                }
                string temp2 = let[j].substr(j2+1, let[j].length()-j2-1);
                if(temp1>temp2){
                    string temp= let[i];
                    let[i] = let[j];
                    let[j] = temp;
                }
                else if (temp1 == temp2){
                    temp1 = let[i].substr(0, j1+1);
                    temp2 = let[j].substr(0, j2+1);
                    if(temp1>temp2){
                        string temp= let[i];
                        let[i] = let[j];
                        let[j] = temp;
                    }
                }
            }    
        }
        
//         for(int i = 0; i < dig.size() - 1; i++){
//             for(int j = i+1; j < dig.size() ; j++){
//                 int j1 = 0;
//                 while(dig[i][j1] != ' '){
//                     j1++;
//                 }
//                 string temp1 = dig[i].substr(j1+1, dig[i].length()-j1-2);
            
//                 int j2 = 0;
//                 while(dig[j][j2] != ' '){
//                     j2++;
//                 }
//                 string temp2 = dig[j].substr(j2+1, dig[j].length()-j2-2);
//                 if(temp1>temp2){
//                     string temp= dig[i];
//                     dig[i] = dig[j];
//                     dig[j] = temp;
//                 }
//             }    
//         }
        
        vector<string> ans;
        for(int i = 0; i < let.size(); i++){
            ans.push_back(let[i]);
        }
        for(int i = 0; i < dig.size(); i++){
            ans.push_back(dig[i]);
        }
        
        return ans;
    }
};
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