<p>You are given a string <code>num</code> consisting of digits only.</p>

<p>Return <em>the <strong>largest palindromic</strong> integer (in the form of a string) that can be formed using digits taken from </em><code>num</code>. It should not contain <strong>leading zeroes</strong>.</p>

<p><strong>Notes:</strong></p>

<ul>
	<li>You do <strong>not</strong> need to use all the digits of <code>num</code>, but you must use <strong>at least</strong> one digit.</li>
	<li>The digits can be reordered.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;444947137&quot;
<strong>Output:</strong> &quot;7449447&quot;
<strong>Explanation:</strong> 
Use the digits &quot;4449477&quot; from &quot;<u><strong>44494</strong></u><u><strong>7</strong></u>13<u><strong>7</strong></u>&quot; to form the palindromic integer &quot;7449447&quot;.
It can be shown that &quot;7449447&quot; is the largest palindromic integer that can be formed.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;00009&quot;
<strong>Output:</strong> &quot;9&quot;
<strong>Explanation:</strong> 
It can be shown that &quot;9&quot; is the largest palindromic integer that can be formed.
Note that the integer returned should not contain leading zeroes.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num.length &lt;= 10<sup>5</sup></code></li>
	<li><code>num</code> consists of digits.</li>
</ul>

class Solution:
    def largestPalindromic(self, num: str) -> str:
        d = {}
        res = []
        for digit in num:
            d[digit] = d.get(digit, 0) + 1
        for key in sorted(d.keys(), reverse=True):
            val = d[key]
            if val == 1: continue # Skip if not enough to form a pair
            if key == '0' and not res: break # Skip if leading 0
            for _ in range(val // 2):
                res.append(key)
            d[key] = val % 2
            if d[key] == 0: del d[key]
        # Edge case: Try to add 0's if we can
        key_0 = '0'
        if key_0 in d and res:
            val = d[key_0]
            for _ in range(val // 2):
                res.append(key_0)
            d[key_0] = d[key_0] % 2
            if d[key_0] == 0: del d[key_0]
        # At the end, squeeze a digit into the middle if we can
        if d:
            middle = [max(d.keys())]
            return "".join(res + middle + res[::-1])
        else:
            return "".join(res + res[::-1])
        
18900