<p>Given two integers <code>left</code> and <code>right</code>, return <em>the <strong>count</strong> of numbers in the <strong>inclusive</strong> range </em><code>[left, right]</code><em> having a <strong>prime number of set bits</strong> in their binary representation</em>.</p>

<p>Recall that the <strong>number of set bits</strong> an integer has is the number of <code>1</code>&#39;s present when written in binary.</p>

<ul>
	<li>For example, <code>21</code> written in binary is <code>10101</code>, which has <code>3</code> set bits.</li>
</ul>

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

<pre>
<strong>Input:</strong> left = 6, right = 10
<strong>Output:</strong> 4
<strong>Explanation:</strong>
6  -&gt; 110 (2 set bits, 2 is prime)
7  -&gt; 111 (3 set bits, 3 is prime)
8  -&gt; 1000 (1 set bit, 1 is not prime)
9  -&gt; 1001 (2 set bits, 2 is prime)
10 -&gt; 1010 (2 set bits, 2 is prime)
4 numbers have a prime number of set bits.
</pre>

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

<pre>
<strong>Input:</strong> left = 10, right = 15
<strong>Output:</strong> 5
<strong>Explanation:</strong>
10 -&gt; 1010 (2 set bits, 2 is prime)
11 -&gt; 1011 (3 set bits, 3 is prime)
12 -&gt; 1100 (2 set bits, 2 is prime)
13 -&gt; 1101 (3 set bits, 3 is prime)
14 -&gt; 1110 (3 set bits, 3 is prime)
15 -&gt; 1111 (4 set bits, 4 is not prime)
5 numbers have a prime number of set bits.
</pre>

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

<ul>
	<li><code>1 &lt;= left &lt;= right &lt;= 10<sup>6</sup></code></li>
	<li><code>0 &lt;= right - left &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int primeCount(string num) {
        int count = 0;
        for (char i : num) {
            if (i == '1')
                count++;
        }
        if (count <= 1) {
            return 0;
        }

        for (int i = 2; i < count; i++) {
            if (count % i == 0) {
                return 0;
            }
        }
        return 1;
    }
    int countPrimeSetBits(int left, int right) {
        int countPrime = 0;

        for (int k = left; k <= right; k++) {
            string num = "";
            int i = 0;
            int binaryNum[32];
            int temp = k;
            while (temp > 0) {
                binaryNum[i] = temp % 2;
                temp = temp / 2;
                i++;
            }
            for (int j = i - 1; j >= 0; j--)
                num+= '0' + binaryNum[j];
                // num += (char)binaryNum[j];
            if(primeCount(num)){
                countPrime++;
            }
        }
        return countPrime;
    }
};
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