<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)))
        n_len = len(digits)

        def count_fewer_digits():
            count = 0
            for l in range(1, n_len):
                if l == 1:
                    count += 9
                else:
                    count += 9 * self.perm(9, l - 1)
            return count

        def count_same_digits():
            seen = set()
            count = 0

            for i in range(n_len):
                for x in range(1 if i == 0 else 0, digits[i]):
                    if x in seen:
                        continue
                    count += self.perm(9 - i, n_len - i - 1)
                if digits[i] in seen:
                    break
                seen.add(digits[i])

            if len(seen) == n_len:
                count += 1

            return count

        total_count = count_fewer_digits() + count_same_digits()
        return total_count

    def perm(self, a, b):
        res = 1
        for i in range(b):
            res *= a - i
        return res


# Driver code to test the solution
solution = Solution()
print(solution.countSpecialNumbers(20))  # Output: 19
print(solution.countSpecialNumbers(5))  # Output: 5
print(solution.countSpecialNumbers(135))  # Output: 110


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