<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 {
public:
    vector<long long> kthPalindrome(vector<int>& queries, int intLength) {
        int subOutRange = pow(10, (intLength - 1) / 2);
        int lastRange = subOutRange * 9;

        int n = queries.size();
        vector<long long> ans(n, 0);
        for(int i = 0; i < n; ++i){
            if(queries[i] > lastRange)
                ans[i] = -1;
            else{
                vector<int> num(intLength, 0);
                int outVal = ceil((double)queries[i] / (double)subOutRange);
                num[0] = outVal;
                num[intLength - 1] = outVal;

                if(intLength > 2) {
                    int remain = queries[i] % subOutRange;
                    if(remain != 0)
                        remain -= 1;
                    else
                        remain = (subOutRange - 1);

                    int mid = (intLength - 1) / 2;
                    while(remain != 0 && mid > 0){
                        num[mid] = remain % 10;
                        num[intLength - mid - 1] = num[mid];

                        remain /= 10;
                        mid -= 1;
                    }
                }

                for(auto it : num)
                    ans[i] = ans[i] * 10 + (long long int)it;
            }
        }

        return ans;
    }
};
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