<p>We call a positive integer <strong>special</strong> if all of its digits are <strong>distinct</strong>.</p>

<p>Given a <strong>positive</strong> integer <code>n</code>, return <em>the number of special integers that belong to the interval </em><code>[1, n]</code>.</p>

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

<pre>
<strong>Input:</strong> n = 20
<strong>Output:</strong> 19
<strong>Explanation:</strong> All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers.
</pre>

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

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong> All the integers from 1 to 5 are special.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 135
<strong>Output:</strong> 110
<strong>Explanation:</strong> There are 110 integers from 1 to 135 that are special.
Some of the integers that are not special are: 22, 114, and 131.</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>9</sup></code></li>
</ul>

class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        def A(m, n):
            return 1 if n == 0 else A(m, n - 1) * (m - n + 1)

        vis = [False] * 10
        ans = 0
        digits = [int(c) for c in str(n)[::-1]]
        m = len(digits)
        for i in range(1, m):
            ans += 9 * A(9, i - 1)
        for i in range(m - 1, -1, -1):
            v = digits[i]
            j = 1 if i == m - 1 else 0
            while j < v:
                if not vis[j]:
                    ans += A(10 - (m - i), i)
                j += 1
            if vis[v]:
                break
            vis[v] = True
            if i == 0:
                ans += 1
        return ans
38