<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>

int dp[100][2][1<<10][2];

class Solution {
public:
    int f(int src,std::string str,int lz,int mask,int ise){
        if(src == str.size()) {
            return 1;
        }
        if(dp[src][lz][mask][ise] != -1) return dp[src][lz][mask][ise];
        int mxx = (ise == 1)?(str[src]-'0') : 9;
        int ans = 0;
        for(int j = 0; j <= mxx; j++){
            if( (lz == 0) && (mask&(1<<j)) ) continue;
            mask |= (1<<j);
            ans += f(src+1,str,lz&(j==0),mask,ise & ( j == mxx) );
            mask ^= (1<<j);
        }
        return dp[src][lz][mask][ise] = ans;
    }
    int numDupDigitsAtMostN(int n) {
        std::string numStr = to_string(n);
        memset(dp,-1,sizeof(dp));
        int ans = (n+1) - f(0,numStr,1,0,1);
        return ans;
    }
};
/// 0 0 0 1
11500