<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)
        return self.dp(0, s, True, 0) - 1

    @lru_cache(None)
    def dp(self, idx, s, limit, used): 
        if idx == len(s): 
            return 1

        ans = 0
        if not limit:
            up = 9
        else:
            up = int(s[idx])
        for i in range(up + 1): 
            if used >> i & 1: 
                continue
            limittt = limit and (i == up)
            if used == 0 and i == 0: 
                ans += self.dp(idx+1, s, limittt, used)
            else:
                ans += self.dp(idx+1, s, limittt, used | 1 << i)
        return ans
266