133 lines
		
	
	
		
			2.5 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			133 lines
		
	
	
		
			2.5 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
---
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id: 5900f38f1000cf542c50fea2
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title: 'Problem 35: Circular primes'
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challengeType: 5
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forumTopicId: 302009
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dashedName: problem-35-circular-primes
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---
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# --description--
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The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
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There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
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How many circular primes are there below `n`, whereas 100 ≤ `n` ≤ 1000000?
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**Note:**
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Circular primes individual rotation can exceed `n`.
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# --hints--
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`circularPrimes(100)` should return a number.
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```js
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assert(typeof circularPrimes(100) === 'number');
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```
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`circularPrimes(100)` should return 13.
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```js
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assert(circularPrimes(100) == 13);
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```
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`circularPrimes(100000)` should return 43.
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```js
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assert(circularPrimes(100000) == 43);
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```
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`circularPrimes(250000)` should return 45.
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```js
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assert(circularPrimes(250000) == 45);
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```
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`circularPrimes(500000)` should return 49.
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```js
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assert(circularPrimes(500000) == 49);
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```
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`circularPrimes(750000)` should return 49.
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```js
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assert(circularPrimes(750000) == 49);
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```
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`circularPrimes(1000000)` should return 55.
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```js
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assert(circularPrimes(1000000) == 55);
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```
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# --seed--
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## --seed-contents--
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```js
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function circularPrimes(n) {
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  return n;
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}
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circularPrimes(1000000);
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```
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# --solutions--
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```js
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function rotate(n) {
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  if (n.length == 1) return n;
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  return n.slice(1) + n[0];
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}
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function circularPrimes(n) {
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  // Nearest n < 10^k
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  const bound = 10 ** Math.ceil(Math.log10(n));
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  const primes = [0, 0, 2];
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  let count = 0;
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  // Making primes array
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  for (let i = 4; i <= bound; i += 2) {
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    primes.push(i - 1);
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    primes.push(0);
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  }
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  // Getting upperbound
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  const upperBound = Math.ceil(Math.sqrt(bound));
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  // Setting other non-prime numbers to 0
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  for (let i = 3; i < upperBound; i += 2) {
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    if (primes[i]) {
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      for (let j = i * i; j < bound; j += i) {
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        primes[j] = 0;
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      }
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    }
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  }
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  // Iterating through the array
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  for (let i = 2; i < n; i++) {
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    if (primes[i]) {
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      let curr = String(primes[i]);
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      let tmp = 1; // tmp variable to hold the no of rotations
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      for (let x = rotate(curr); x != curr; x = rotate(x)) {
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        if (x > n && primes[x]) {
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          continue;
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        }
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        else if (!primes[x]) {
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          // If the rotated value is 0 then it isn't a circular prime, break the loop
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          tmp = 0;
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          break;
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        }
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        tmp++;
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        primes[x] = 0;
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      }
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      count += tmp;
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    }
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  }
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  return count;
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}
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```
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