<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) {
        string result="";
        int n = s.length();

        int a=0, b=0;

        for(int i=0;i<n;i++)
        {
            if(s[i] >='a' && s[i] <= 'z')
                a++;
            
            if(s[i] >='0' && s[i] <= '9')
                b++;
        }

        queue<char> s1;
        queue<char> s2;

        for(int i=0;i<n;i++)
        {
            if(s[i] >='a' && s[i] <= 'z')
                s1.push(s[i]);
            
            else
                s2.push(s[i]);
        }

        if((a == b) || (a-1 == b) ||(a == b-1))
        {
            if(a<b)
            {
                queue<char> temp = s1;
                s1 = s2;
                s2 = temp;
            }

            for(int i=0;i<n;i++)
            {
                if(i%2==0)
                {
                    result += s1.front();
                    s1.pop();
                }
            
                else
                {
                    result += s2.front();
                    s2.pop();
                }
            }
        }
        return result;
    }
};
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