<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) {
        
        unordered_map<int,int> mp;
        mp[0]=  0;
        mp[1]=  1;
        mp[8]=  8;
        mp[2] = 5;
        mp[5] = 2;
        mp[6] = 9;
        mp[9] = 6;
        int ans =0;
        for(int i=1;i<=n;i++){

            int num = i;
            bool flag = true;
            int nn=0;
            int cnt =0;
            while(num){
                
                int rem = num%10;
                if(mp.find(rem)==mp.end()){
                    flag = false;
                    break;
                }
                else{
                    nn = pow(10,cnt)*mp[rem]+nn;
                }
                cnt++;
                num/=10;
            }
            if(nn==i ) flag = false;
            
            if(flag)
            ans++;
        }
        return ans;
    }
};
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