<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[2][2][100][1<<10];
    int f(bool tight, bool leading, int n, int bits, string num) {
        int ub = num[num.size() - n] - '0';
        if (tight == 0) {
            ub = 9;
        }
        if (n == 0) {
            return 1;
        }
        if(dp[tight][leading][n][bits]!=-1){
            return dp[tight][leading][n][bits];
        }
        int ans = 0;
        for (int i = 0; i <= ub; i++) {
            if (((bits >> i) & 1) && leading==0) {
                continue;
            }
            if(leading && i==0){
                ans+=f(tight&(i==ub),true,n-1,bits,num);
            }
            else{
                ans+=f(tight&(i==ub),false,n-1,bits|(1<<i),num);
            }
        }
        return dp[tight][leading][n][bits]=ans;
    }
    int numDupDigitsAtMostN(int n) {
        memset(dp,-1,sizeof(dp));
        string num = to_string(n);
        return n-f(1, 1, num.size(), 0, num)+1 ;
    }
};
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