Reformat The String
<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>
String
class Solution {
public:
    string reformat(string s) {
        string a = "", b = "";
        for (char c : s) {
            if (isdigit(c))
                a += c;
            else
                b += c;
        }
        int m = a.size(), n = b.size();
        if (abs(m - n) > 1) return "";
        string ans = "";
        for (int i = 0; i < min(m, n); ++i) {
            if (m > n) {
                ans += a[i];
                ans += b[i];
            } else {
                ans += b[i];
                ans += a[i];
            }
        }
        if (m > n) ans += a[m - 1];
        if (m < n) ans += b[n - 1];
        return ans;
    }
};