<p>Given two integers <code>n</code> and <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>lexicographically smallest integer in the range</em> <code>[1, n]</code>.</p>

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

<pre>
<strong>Input:</strong> n = 13, k = 2
<strong>Output:</strong> 10
<strong>Explanation:</strong> 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><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, k = 1
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution(object):
    def findKthNumber(self, n, k):
        curr = 1
        k -= 1
        
        while k > 0:
            step = self.countSteps(n, curr, curr + 1)
            # If the steps are less than or equal to k, we skip this prefix's subtree
            if step <= k:
                # Move to the next prefix and decrease k by the number of steps we skip
                curr += 1
                k -= step
            else:
                # Move to the next level of the tree and decrement k by 1
                curr *= 10
                k -= 1

        return curr

    # To count how many numbers exist between prefix1 and prefix2
    def countSteps(self, n, prefix1, prefix2):
        steps = 0
        while prefix1 <= n:
            steps += min(n + 1, prefix2) - prefix1
            prefix1 *= 10
            prefix2 *= 10
        return steps
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