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