<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[10][2][1025][2];

    int count(int idx,bool tight,int bitmask,bool isrepeated,string&num){
        if(idx==num.size()){
            return (isrepeated&&bitmask)?1:0;
        }
        if(dp[idx][tight][bitmask][isrepeated]!=-1) return dp[idx][tight][bitmask][isrepeated];
        int lo=0;
        int hi=tight?(num[idx]-'0'):9;
        int cnt=0;
        if(!bitmask){
            cnt+=count(idx+1,false,bitmask,isrepeated,num);
        }
        for(int i=lo;i<=hi;i++){
            if(!bitmask && i==0) continue;
            int repeated=bitmask & (1<<i);
            if(repeated){
                cnt+=count(idx+1,tight&&(i==hi),bitmask,true,num);
            }
            else{
                cnt+=count(idx+1,tight&&(i==hi),bitmask|(1<<i),isrepeated,num);
            }
        }
        return dp[idx][tight][bitmask][isrepeated]=cnt;
    }
    int numDupDigitsAtMostN(int n) {
        memset(dp,-1,sizeof(dp));
        string num = to_string(n);
        return count(0,true,0,false,num);
    }
};
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