Count Special Integers
<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>
Math; Dynamic Programming
class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        @cache
        def dfs(i: int, mask: int, lead: bool, limit: bool) -> int:
            if i >= len(s):
                return int(lead ^ 1)
            up = int(s[i]) if limit else 9
            ans = 0
            for j in range(up + 1):
                if mask >> j & 1:
                    continue
                if lead and j == 0:
                    ans += dfs(i + 1, mask, True, limit and j == up)
                else:
                    ans += dfs(i + 1, mask | 1 << j, False, limit and j == up)
            return ans

        s = str(n)
        return dfs(0, 0, True, True)
