<p>An integer <code>x</code> is a <strong>good</strong> if after rotating each digit individually by 180 degrees, we get a valid number that is different from <code>x</code>. Each digit must be rotated - we cannot choose to leave it alone.</p>

<p>A number is valid if each digit remains a digit after rotation. For example:</p>

<ul>
	<li><code>0</code>, <code>1</code>, and <code>8</code> rotate to themselves,</li>
	<li><code>2</code> and <code>5</code> rotate to each other (in this case they are rotated in a different direction, in other words, <code>2</code> or <code>5</code> gets mirrored),</li>
	<li><code>6</code> and <code>9</code> rotate to each other, and</li>
	<li>the rest of the numbers do not rotate to any other number and become invalid.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the number of <strong>good</strong> integers in the range </em><code>[1, n]</code>.</p>

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

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int rotatedDigits(int n) {
        vector< int> dp1( 10001, 0);
        vector< int> dp2( 10001, 0);

        set< int> st1 = {0,1,8}, st2 = {2,5,6,9};

        for( int i : st1){
            dp2[ i] = 1;
        }

        for( int i : st2){
            dp1[ i] = 1;
        }

        for( int i = 10; i <= n; i++){
            int num = i;

            int d = num % 10;
            num = num /10;

            if( st2.find( d) != st2.end() && ( dp1[ num] || dp2[ num])){
                dp1[ i] = 1;
            }else if( st1.find( d) != st1.end() && dp2[ num]){
                dp2[ i] = 1;
            }else if( st1.find( d) != st1.end() && dp1[ num]){
                dp1[ i] = 1;
            }
        }

        int ans = 0;
        for( int i = 1; i <= n; i++){
            ans += dp1[ i];
        }

        return ans;
    }
};
12400