<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 rev(int num){
        int nn = 0;
        while(num>0){
            int rem = num%10;
            nn = nn*10 + rem;
            num = num/10;
        }
        return nn;
    }

    int rotatedDigits(int n) {
        unordered_map<int, int> m;
        m[1] = 1;
        m[0] = 0;
        m[8] = 8;
        m[2] = 5;
        m[5] = 2;
        m[6] = 9;
        m[9] = 6;

        int count = 0;
        for(int i=0; i<=n; i++){
            int num = i;
            int nn = 0;
            if(num == 0) continue;
            bool flag = true;
            int p = 0;
            while(num>0){
                int dig = num%10;
                if(m.find(dig) == m.end()){
                    flag = false;
                    break;
                }
                else{
                    nn = m[dig]*pow(10, p) + nn;
                }
                num = num/10;
                p++;
            }
            // cout<<nn<<" "<<i<<endl;
            if(flag && nn!=i){
                count++;
            }
        }

        return count;

    }
};
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