<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 {
public:
int dp[1<<10][10][2][2];
    int numDupDigitsAtMostN(int n) {
        memset(dp, -1, sizeof(dp));
        string s = to_string(n);
        int mask = 0;
        return n-f(s, 0, mask, 1, 1)+1;
    }
    int f(string s, int idx, int mask, int lz, int bound){
        if(idx == s.size()) return 1;
        if(dp[mask][idx][lz][bound] != -1) return dp[mask][idx][lz][bound];
        int mx = bound? s[idx]-'0' : 9;
        int cnt = 0;
        for(int j=0; j<=mx; j++){
            if((mask&(1<<j)) && (lz==0)) continue;
            mask |= 1<<j;
            cnt += f(s, idx+1, mask, lz&(j==0), bound&(j==mx));
            mask ^= 1<<j;
        }
        return dp[mask][idx][lz][bound] = cnt;
    }
};
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