<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:
        odd = False
        odd_val = ""
        arr = [0] * 10
        for item in num:
            arr[int(item)] += 1
        if arr[0] == sum(arr):
            return "0"
        def fn(arr):
            maxx = 0
            found = False
            for i in range(len(arr) - 1, 0, -1):
                if arr[i] == 1 and found == False:
                    found = True
                    maxx = str(i)
                if arr[i] > 1:
                    return False
            return maxx

        if fn(arr) == False:
            front = ""
            back = ""
            for i in range(len(arr) - 1, -1, -1):
                if arr[i] % 2 != 0 and odd == False:
                    odd_val = str(i)
                    odd = True
                front += str(i) * (arr[i] // 2)
                back += str(i) * (arr[i] // 2)
            return(front + odd_val + back[::-1])
        else:
            return(fn(arr))

                    

                        
                    
                
                
18000