<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<char> nums;
        vector<char> chars;

        for (int i = 0; i < s.length(); i++) {
            if (isalpha(s[i]))
                chars.push_back(s[i]);
            if (isdigit(s[i]))
                nums.push_back(s[i]);
        }

        if (abs(static_cast<int>(nums.size()) - static_cast<int>(chars.size())) > 1)
            return "";

        string answer = "";
        if (nums.size() >= chars.size()) {
            int i = 0, j = 0;
            while (i < nums.size() || j < chars.size()) {
                if ((i + j) % 2 == 0) {
                    answer = answer + nums[i];
                    i++;
                } else {
                    answer = answer + chars[j];
                    j++;
                }
            }
        }
        else
        {
            int i = 0, j = 0;
            while (i < chars.size() || j < nums.size()) {
                if ((i + j) % 2 == 0) {
                    answer = answer + chars[i];
                    i++;
                } else {
                    answer = answer + nums[j];
                    j++;
                }
            }
        }

        return answer;
    }
};
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