<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<stack<char>> sts {{},{}};

        for(char& c : s) {
            if(isdigit(c)) {
                sts[0].push(c);
            }
            if(isalpha(c)) {
                sts[1].push(c);
            }
        }

        int d = sts[0].size();
        int a = sts[1].size();

        if(abs(d-a)> 1) {
            return "";
        }

        int m ;
        int off;
        if(d > a) {
            m = d;
            off = 0;
        } else {
            m = a;
            off = 1;
        }
        string ret;

        for(int i = 0 ; i < m*2 ; i ++) {
            if(!sts[(i+off)%2].empty()) {
                char c = sts[(i+off)%2].top();
                sts[(i+off)%2].pop();
                ret.append(1,c);
            }
        }

        return ret;
    }
};
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