<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:
    void digit(vector<int>& num, int k) {   
        for(int i = 0; i < 4; i++) {
            num.push_back(k % 10);
            k = k / 10;
            if(k == 0) {
                return;
            }
        }
    }


    int rotatedDigits(int n) {

        set<int> s;
        s.insert(1);
        s.insert(2);
        s.insert(5);
        s.insert(6);
        s.insert(8);
        s.insert(9);
        s.insert(0);

        set<int> v;
        v.insert(0);
         v.insert(8);
         v.insert(1);

        int count = 0;
        for(int i = 1 ; i <= n ; i++)
        {
            vector<int> num;
            digit(num,i);
            int t = 0;
            bool valid = true;
            for(int j = 0 ; j < num.size() ; ++j)
            {   
                if(s.find(num[j]) == s.end())
                {   
                    valid = false;
                    break;
                }
                else
                {
                    if(v.find(num[j]) == v.end())
                    {
                        t = 1;
                    }
                }
               
            }
            if(valid && t == 1)
            {
                count++;
            }

        }
    return count;
    }
};
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