<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:
        digits = list(map(int, str(n + 1)))
        length = len(digits)
        special_count = 0
        
        # Count all numbers with fewer digits
        for i in range(1, length):
            special_count += 9 * self.perm(9, i - 1)
        
        # Count numbers with the same number of digits but less than n
        seen = set()
        for i in range(length):
            for j in range(1 if i == 0 else 0, digits[i]):
                if j not in seen:
                    special_count += self.perm(9 - i, length - i - 1)
            if digits[i] in seen:
                break
            seen.add(digits[i])
        
        return special_count
    
    def perm(self, m, n):
        result = 1
        for i in range(n):
            result *= (m - i)
        return result
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