<p>Given an integer <code>n</code>, return <em>the number of positive integers in the range </em><code>[1, n]</code><em> that have <strong>at least one</strong> repeated digit</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 20
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only positive number (&lt;= 20) with at least 1 repeated digit is 11.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 100
<strong>Output:</strong> 10
<strong>Explanation:</strong> The positive numbers (&lt;= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 1000
<strong>Output:</strong> 262
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def numDupDigitsAtMostN(self, n: int) -> int:
        cache={}
        s=str(n)
        def dfs(index,isUpperLimit,hasLeadingZeroes,mask):
            if index==len(s):
                if hasLeadingZeroes:
                    return 0
                return 1
            if (index,isUpperLimit,hasLeadingZeroes,mask) in cache:
                return cache[(index,isUpperLimit,hasLeadingZeroes,mask)]
            res=0
            if hasLeadingZeroes:
                res+=dfs(index+1,False,True,mask)
            low=1 if hasLeadingZeroes else 0
            high=int(s[index]) if isUpperLimit else 9

            for i in range(low,high+1):
                if not mask & 1<<i:
                    res+=dfs(index+1,True if isUpperLimit and i==high else False,False,mask | 1<<i)
            
            cache[(index,isUpperLimit,hasLeadingZeroes,mask)]=res

            return res
        
        return n-dfs(0,True,True,0)
20900