<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>

#define ll long long
class Solution {
public:
    int rotatedDigits(int n) {
        vector<int> v(10,0);
        int dp[10005];
        for(int i=0;i<10;i++){
            if(i==0 || i==1 || i==8) dp[i] = 0;
            else if(i==2 || i==5 || i==6 || i== 9) dp[i]=1;
            else dp[i]=-1;
        }
        v[0]=dp[0];
        for(int i=1;i<=9;i++){
            if(dp[i]!=-1){
                v[i] = v[i-1]+dp[i];
            }
            else{
                v[i] = v[i-1];
            }
        }
        if(n<=9){
            return v[n];
        }
        ll ans = 4;
        for(int i=11;i<=n;i++){
            int q = i/10;
            int r = i%10;
            if(dp[q]==-1 || dp[r]==-1){
                dp[i] = -1;
            }
            else if(dp[q]==1 || dp[r]==1){
                dp[i]=1;
                ans++;
            }
            else{
                dp[i] = 0;
            }
        }
        return ans;
    }
};
8100