<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>

class Solution {
public:
    string reformat(string s) {
        vector<int> cnt(36, 0);
        int c = 0;
        for (int i = 0; i < s.size(); i++) {
            if (isalpha(s.at(i))) { cnt.at(s.at(i) - 'a')++; c++; }
            else cnt.at(s.at(i) - '0' + 26)++;
        }
        int i = s.size() - 2 * c;
        if (abs(i) > 1) return "";
        i = 0;
        int j = 26;
        c = (c < s.size() - c);
        s.clear();
        if (c) {
            while (j < 36 && !cnt.at(j)) j++;
            s.push_back('0' + j - 26);
            cnt.at(j)--;
        }
        do {
            while (i < 26 && !cnt.at(i)) i++;
            if (i < 26) {
                s.push_back('a' + i);
                cnt.at(i)--;
            }
            while (j < 36 && !cnt.at(j)) j++;
            if (j < 36) {
                s.push_back('0' + j - 26);
                cnt.at(j)--;
            }
        } while (i < 26 || j < 36);
        return s;
    }
};
8900