<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], intLength: int) -> List[int]:
        # think the palindromes in half
        # e.g. len  = 4 we only consider the first 2 digits
        # half: 10, 11, 12, 13, 14, ..., 19, 20, 
        # full: 1001, 1111, 1221, 1331, ...
        # e.g. len = 5 we consider the first 3 digits
        # half: 100, 101, 102, ...
        # full: 10001, 10101, 10201, ...
        
        result = []
        
        for i in queries:
            result.append(self.generatePalindrome(intLength, i))
        
        return result
    
    def generatePalindrome(self, length, num):
        # index start from 0
		# e.g. num =1 means we want to find the most smallest palindrome, then its index is 0
		# e.g. num =2 means we want to find the second most smallest palindrome, then its index is 1
        index = num -1
        
		# if the length is even
		# we only think about the fisrt half of digits
        if length % 2 == 0:
            cur = int('1' + '0' * (length // 2 -1))
            maxLength = len(str(cur))
            cur += index
            
            if len(str(cur)) > maxLength:
                return -1
            
            else:
                cur = str(cur)
                cur = cur + cur[::-1]
                cur = int(cur)
                return cur
				
        # if the length is odd
		# we consider first (length // 2 + 1) digits
        else:
            cur = int('1' + '0' * (length // 2))
            maxLength = len(str(cur))
            cur += index
            
            if len(str(cur)) > maxLength:
                return -1
            
            else:
                cur = str(cur)
                temp = str(cur)[:-1]
                cur = cur + temp[::-1]
                cur = int(cur)
                return cur
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