<p>Given an integer array <code>queries</code> and a <strong>positive</strong> integer <code>intLength</code>, return <em>an array</em> <code>answer</code> <em>where</em> <code>answer[i]</code> <em>is either the </em><code>queries[i]<sup>th</sup></code> <em>smallest <strong>positive palindrome</strong> of length</em> <code>intLength</code> <em>or</em> <code>-1</code><em> if no such palindrome exists</em>.</p>

<p>A <strong>palindrome</strong> is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros.</p>

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

<pre>
<strong>Input:</strong> queries = [1,2,3,4,5,90], intLength = 3
<strong>Output:</strong> [101,111,121,131,141,999]
<strong>Explanation:</strong>
The first few palindromes of length 3 are:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...
The 90<sup>th</sup> palindrome of length 3 is 999.
</pre>

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

<pre>
<strong>Input:</strong> queries = [2,4,6], intLength = 4
<strong>Output:</strong> [1111,1331,1551]
<strong>Explanation:</strong>
The first six palindromes of length 4 are:
1001, 1111, 1221, 1331, 1441, and 1551.
</pre>

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

<ul>
	<li><code>1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= queries[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= intLength&nbsp;&lt;= 15</code></li>
</ul>

class Solution:
    def kthPalindrome(self, queries: List[int], n: int) -> List[int]:
        
        if n == 1:
            res = []

            for q in queries:
                if q > 9:
                    res.append(-1)
                else:
                    res.append(q)
            
            return res
        
        queries = [(i, v) for i, v in enumerate(queries)]
        queries.sort(key = lambda x: x[1])

        # Even case
        
        half_len = n // 2
        left = '1' + ('0' * (half_len - 1))
        
        if n & 1:
            left += '0'
        
        curr_rank = 1

     
        

        def get_next_pali(left, diff = 1):
            return str(int(left) + diff)
        
        res = [-1] * len(queries)
        fail = False
        if n & 1 == 0:
                    curr_pali = left + left[::-1]
        else:
                    curr_pali = left + left[:len(left) - 1][::-1]
 
        

        for idx, k in queries:

            if curr_rank < k:
                left = get_next_pali(left, k - curr_rank)

                curr_rank = k

            # while curr_rank < k:
                # left = get_next_pali(left)
            
            

            if n & 1 == 0:
                curr_pali = left + left[::-1]
            else:
                curr_pali = left + left[:len(left) - 1][::-1]

            if len(curr_pali) > n:
                fail = True
                break
            
        
        
            if curr_rank == k:
                res[idx] = int(curr_pali)

        return res            

    

503