<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:
        vals = list(map(int, str(n)))
        
        @cache
        def fn(i, m, on): 
            """Return count at index i with mask m and profile flag (True/False)"""
            ans = 0 
            if i == len(vals): return 1
            for v in range(vals[i] if on else 10 ): 
                if m & 1<<v == 0: 
                    if m or v: ans += fn(i+1, m ^ 1<<v, False)
                    else: ans += fn(i+1, m, False)
            if on and m & 1<<vals[i] == 0: ans += fn(i+1, m ^ 1<<vals[i], True)
            return ans 
        
        return fn(0, 0, True)-1
208