/**
<p>Given integers <code>n</code> and <code>k</code>, find the lexicographically k-th smallest integer in the range from <code>1</code> to <code>n</code>.</p>

<p>Note: 1 &le; k &le; n &le; 10<sup>9</sup>.</p>

<p><b>Example:</b>
<pre>
<b>Input:</b>
n: 13   k: 2

<b>Output:</b>
10

<b>Explanation:</b>
The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.
</pre>
</p>
<p>给定整数&nbsp;<code>n</code>&nbsp;和&nbsp;<code>k</code>，找到&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;中字典序第&nbsp;<code>k</code>&nbsp;小的数字。</p>

<p>注意：1 &le; k &le; n &le; 10<sup>9</sup>。</p>

<p><strong>示例 :</strong></p>

<pre>
<strong>输入:</strong>
n: 13   k: 2

<strong>输出:</strong>
10

<strong>解释:</strong>
字典序的排列是 [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]，所以第二小的数字是 10。
</pre>
<p>给定整数&nbsp;<code>n</code>&nbsp;和&nbsp;<code>k</code>，找到&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;中字典序第&nbsp;<code>k</code>&nbsp;小的数字。</p>

<p>注意：1 &le; k &le; n &le; 10<sup>9</sup>。</p>

<p><strong>示例 :</strong></p>

<pre>
<strong>输入:</strong>
n: 13   k: 2

<strong>输出:</strong>
10

<strong>解释:</strong>
字典序的排列是 [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]，所以第二小的数字是 10。
</pre>
**/


class Solution {
    fun findKthNumber(n: Int, k: Int): Int {
        
    }
}
/**
<p>Given integers <code>n</code> and <code>k</code>, find the lexicographically k-th smallest integer in the range from <code>1</code> to <code>n</code>.</p>

<p>Note: 1 &le; k &le; n &le; 10<sup>9</sup>.</p>

<p><b>Example:</b>
<pre>
<b>Input:</b>
n: 13   k: 2

<b>Output:</b>
10

<b>Explanation:</b>
The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.
</pre>
</p>
<p>给定整数&nbsp;<code>n</code>&nbsp;和&nbsp;<code>k</code>，找到&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;中字典序第&nbsp;<code>k</code>&nbsp;小的数字。</p>

<p>注意：1 &le; k &le; n &le; 10<sup>9</sup>。</p>

<p><strong>示例 :</strong></p>

<pre>
<strong>输入:</strong>
n: 13   k: 2

<strong>输出:</strong>
10

<strong>解释:</strong>
字典序的排列是 [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]，所以第二小的数字是 10。
</pre>
<p>给定整数&nbsp;<code>n</code>&nbsp;和&nbsp;<code>k</code>，找到&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;中字典序第&nbsp;<code>k</code>&nbsp;小的数字。</p>

<p>注意：1 &le; k &le; n &le; 10<sup>9</sup>。</p>

<p><strong>示例 :</strong></p>

<pre>
<strong>输入:</strong>
n: 13   k: 2

<strong>输出:</strong>
10

<strong>解释:</strong>
字典序的排列是 [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]，所以第二小的数字是 10。
</pre>
**/


class Solution {
    fun findKthNumber(n: Int, k: Int): Int {
        
    }
}