<p>You are given an alphanumeric string <code>s</code>. (<strong>Alphanumeric string</strong> is a string consisting of lowercase English letters and digits).</p>

<p>You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.</p>

<p>Return <em>the reformatted string</em> or return <strong>an empty string</strong> if it is impossible to reformat the string.</p>

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

<pre>
<strong>Input:</strong> s = &quot;a0b1c2&quot;
<strong>Output:</strong> &quot;0a1b2c&quot;
<strong>Explanation:</strong> No two adjacent characters have the same type in &quot;0a1b2c&quot;. &quot;a0b1c2&quot;, &quot;0a1b2c&quot;, &quot;0c2a1b&quot; are also valid permutations.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> &quot;leetcode&quot; has only characters so we cannot separate them by digits.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1229857369&quot;
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> &quot;1229857369&quot; has only digits so we cannot separate them by characters.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of only lowercase English letters and/or digits.</li>
</ul>

#include<bits/stdc++.h>
class Solution {
public:
    string reformat(string s) {
        vector<int>l;
        vector<int>d;
         string res = "";
         int a=0,b=0;
        for(int i=0;i<s.size();i++){
            if(isalpha(s[i])){
                l.push_back(s[i]);
            }
             else{
                d.push_back(s[i]);
            }
        }
        sort(l.begin(),l.end());
        sort(d.begin(),d.end());
        if (abs((int)l.size() - (int)d.size()) > 1) {
            return "";
        }

        if (l.size() > d.size()) {
            std::swap(l, d);
        }
        while(a < l.size() && b<d.size()){
             res += d[b++];
            res += l[a++];
         }
                 if (b < d.size()) {
            res += d[b++];
        }

        return res;
    }
};
10800