[LeetCode Sync] Runtime - 21 ms (97.65%), Memory - 19.4 MB (64.02%)

Author: NamVr

my-folder/0767-prime-number-of-set-bits-in-binary-representation/README.md

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+<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>

my-folder/0767-prime-number-of-set-bits-in-binary-representation/solution.py

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+class Solution:
+    def countPrimeSetBits(self, left: int, right: int) -> int:
+        N = 22
+        sieve = [True] * N
+        sieve[0] = sieve[1] = False
+
+        i = 2
+        while i*i < N:
+            if sieve[i] < N:
+                for j in range(i*i, N, i):
+                    sieve[j] = False
+            i += 1
+        
+        res = 0
+
+        for i in range(left, right+1):
+            #if sieve[bin(i)[2:].count('1')]:
+            if sieve[i.bit_count()]:
+                res += 1
+
+        return res