<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:
        s = str(n)
        d = len(s); ans = 0

        for i in range(1, d):
            x, k = 1, 9
            for _ in range(i - 1):
                x *= k; k -= 1
            ans += 9 * x
        
        used = [0] * 10
        for i in range(d):
            sm = 0
            for j in range(ord(s[i]) - ord('0')):
                if not used[j]: sm += 1

            if i == 0: sm -= 1

            nxt = 1; r = 10 - i - 1
            for j in range(i + 1, d):
                nxt *= r; r -= 1
            ans += sm * nxt

            if not used[ord(s[i]) - ord('0')]:
                used[ord(s[i]) - ord('0')] = 1
            else:
                break

        if sum(used) == d:
            ans += 1
        return ans


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