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function primeFinder(n) {
const arr = [];
for (let i = 2; i <= n; i++) {
let isPrime = true;
let root = Math.floor(Math.sqrt(i));
for (let j = 2; j <= root; j++) {
if (i % j === 0) {
isPrime = false;
break;
}
}
if (isPrime) {
arr.push(i);
}
}
return arr;
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
function primeFinder(n) {
let primeNumbers = [];
for (let i = 1; i <= n; i++) {
if (isItPrime(i)) {
primeNumbers.push(i);
}
}
return primeNumbers;
}
function isItPrime(n) {
if (n === 1) {
return false;
} else if (n === 2) {
return true;
}
for (let i = n - 1; i > 1; i--) {
if (n % i === 0) {
return false;
}
}
return true;
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
I’m not a big fan of nested loops, too much room for confusion - but here it is anyways
function primeFinder(n) {
// Write your code here
let result = []
for(let i = 1; i <= n; i++) {
let count = 0
for(let j = 2; j <= i; j++) {
if(Number.isInteger(i/j)) {
count++
}
}
if(count === 1) {
result.push(i)
}
}
return result
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
function primeFinder(n) {
const isPrime = function (p) {
if (p < 2) {
return false
}
if (p > 2) {
let x = 2
while (x <= Math.sqrt(p)) {
if (p % x == 0) {
return false
}
x += x % 2 ? 2 : 1
}
}
return true
}
const primes = []
for (let i = 2; i <= n; i++) {
if (isPrime(i)) {
primes.push(i)
}
}
return primes
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
Little bit clumsy but it can be cleaned up on refactor.
It follows that since we are building an array of primes, it could be used in the isPrime() function…
Taking a clue from yours, I came up with this, but still have to see if it passes the tests…
const primeFinder = function (n) {
const primes = [false, false].concat(new Array(n - 1).fill(true));
const r = Math.floor(Math.sqrt(n) + 1);
for (let y = 2; y < r; y++) {
for (let x = 2 * y; x < n; x += y) {
primes[x] = false;
}
}
primes.forEach((x, i) => {
if (x) {
primes[i] = i;
}
})
return primes.filter(x => !! x);
}
console.log(primeFinder(103))
BRB
Had to make a suspected tweak…
const primeFinder = function (n) {
const primes = [false, false].concat(new Array(n - 1).fill(true));
const r = Math.floor(Math.sqrt(n) + 1);
for (let y = 2; y < r; y++) {
for (let x = 2 * y; x <= n; x += y) {
primes[x] = false;
}
}
primes.forEach((x, i) => {
if (x) {
primes[i] = i;
}
})
return primes.filter(x => !! x);
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
Big-O wise, when we see nested loops it’s automatically N^2 but in this case the first loop is only iterating over the square root of n, and the second is leaping from point to point with lessening frequency. There never truly is an N-squared scenario. Still, worst case. Hmmm.
Interesting thing to note, boolean type uses four bytes, number type uses eight. For anyone wondering if 0 and 1 could be used for brevity.
Shouldn’t be whining; common multiples are getting overwritten steadily. That’s some work.
Mental logic. This problem doesn’t solve itself. There are edge cases that we as the programmer must consider and explore; at the least, ponder. Toss in the forEach and that makes five mental logic considerations that the code could not solve by itself, things that the intuitive algorithm required to align itself. This is not something we can take lightly. Equip your thinking for this type of case, plus or minus one, etc.
// Return true if n is prime
function isPrime(n)
{
if (n < 2) return false;
for (let i = 2; i <= Math.sqrt(n); i++) {
if (n % i === 0) return false
}
return true;
}
function primeFinder(n) {
let result = [];
for (let i = 0; i <= n; i++) {
if (isPrime(i)) result.push(i)
}
return result;
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
So I thought I’d try to code out some of the simpler cases out right off the bat instead of looping through those numbers, but I wonder if this would be able to handle larger cases. Am I just wasting my time with these additional actions?
function primeFinder(n) {
// Write your code here
const commonDiv = [2, 3, 5, 7];
// Validate parameter > 1
if (n === 1) {
return null;
}
// Get initial list
const prime = commonDiv.filter(number => number <= n);
// Confirm if paramater > 10
if (n <= 10) {
return prime;
}
for (let i = 11; i <= n; i++) {
if (!commonDiv.some(number => i % number === 0)) {
prime.push(i);
}
}
return prime;
}
console.log(primeFinder(50))
// Leave this so we can test your code:
module.exports = primeFinder;
I also tried making a reusable prime number ID’er function which would be used to populate an array. I originally made the function check if ‘n’ was itself actually a prime number (why use resources if it isn’t?), but that seemed to complicate the testing. Let me know what you think.
const primeFinder = n => {
let primes = [];
const isPrime = n => {
let divisors = 0;
for (let i=1; i<=n; i++) {
if (n % i === 0) {
divisors++;
}
}
if (divisors === 2) {
return true;
} else {
return false;
}
}
for (let i=2; i<=n; i++) {
if (isPrime(i)) {
primes.push(i)
};
}
return primes;
} // Have a nice day
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
Tried this challenge with while loops. It is a bit longer than other solutions but it gets the job done, so i’m still proud.
function primeFinder(n) {
// Input primary check. No numbers below 2 allowed.
if (n < 2) {
return 'There are no prime numbers below 2.'
}
const primeNumbArray = [
2
];
let counter = primeNumbArray[primeNumbArray.length - 1];
while (counter <= n) {
let index = 0;
let loopSwitch = true;
// Test: Can "counter" be divided with a 0 remainder by any number in "primeNumbArray"?
while (loopSwitch) {
if (!primeNumbArray[index]) {
// no: "counter" is a prime number
primeNumbArray.push(counter)
}
if (primeNumbArray[index] && counter % primeNumbArray[index]) {
index++
} else {
// yes: "counter" is a multiple of a prime number already stored.
counter++;
loopSwitch = false
}
}
}
return primeNumbArray
}
console.log(primeFinder(100));
// Leave this so we can test your code:
module.exports = primeFinder;
Edit: after watching other solutions i realized my code had some redundant body fat that could be removed. There’s always room for improving.
function primeFinder(n) {
if (typeof n !== 'number' || Number.isNaN(n)) {
return 'Please insert a number.'
}
if (n < 2) {
return 'There are no prime numbers below 2.'
}
const primeNumbs = [];
function isPrime(num) {
for (let i = 0; i < primeNumbs.length; i++) {
if (num % primeNumbs[i] === 0) {
return false
}
}
return true
}
for (let counter = 2; counter <= n; counter++) {
if (isPrime(counter)) {
primeNumbs.push(counter)
}
}
return primeNumbs
}
console.log(primeFinder(11));
// Leave this so we can test your code:
module.exports = primeFinder;
function primeFinder(n) {
let array = [];
for (let i = 2; i <= n; i++) {
array.push(i);
}
let prime = array.filter((n) => {
if (
n === 2 ||
n === 3 ||
n === 5 ||
n === 7 ||
(n / n === 1 &&
n % 2 !== 0 &&
n % 3 !== 0 &&
n % 5 !== 0 &&
n % 7 !== 0 )
)
return n;
});
return prime;
}
console.log(primeFinder(100))
function primeFinder(n) {
let primes = [];
for (let i = 1; i <= n; i++) {
let factors = [];
// Find factors of i
for (let j = 1; j <= i; j++) {
// If no remainder then factor
if (i % j === 0) {
factors.push(j);
}
}
// Prime numbers only have 2 factors - 1, p
if (factors.length === 2) {
// If number has 2 factors one is 1 if other is p then prime
if (factors.includes(i)) {
primes.push(i);
}
}
}
return primes;
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
function primeFinder(n) {
let primes = [];
for (let i = 2; i <= n; i++) {
if (isPrime(i, 2)) {
primes.push(i)
}
}
return primes;
}
// isPrime takes two integers and returns a Boolean. p is the target int and n is the iterator seeded as 2, the smallest prime number
function isPrime(p, n) {
// if 0 or 1 then not prime
if (p == 0 || p == 1) {
return false;
}
// if p is equal to iterator then I have exhausted all the possibilities and its prime
if (p == n) {
return true;
}
// if the p mod n is zero then not prime
if (p%n == 0) {
return false
}
// recursive function: increment the iterator and try again until true or false
return isPrime(p, n+1);
}
console.log(primeFinder(11));
I reuse this isPrime function for all my prime number challenges. I can’t remember where I found it but I translated it from python into go and now javascript.
function primeFinder(n) {
// Write your code here
let results = new Array();
//test each number, i, from 2 to n
for (let i = 2; i <= n; i++)
{
let prime = true;
//test from 2 to i
for (let j = 2; j < i; j++)
{
if (i%j == 0)
{
prime = false;
}
}
if (prime == true)
{
results.push(i);
}
}
return results;
}
// Leave this so we can test your code:
module.exports = primeFinder;
This is my solution.
First I create a function to check if it is prime or no.
So, I aply this function in any number.
function primeFinder(n) {
// Write your code here
const prime_list = []
function check(number){
if(number < 2){
return false
}
for (let i = 2; i < number; i++) {
if (number%i === 0){
return false
}
}
return true
}
for (let i = 0; i <= n; i++) {
if (check(i)){
prime_list.push(i)
}
}
return prime_list
}
console.log(primeFinder(11))
// Leave this so we can test your code:
module.exports = primeFinder;
) below.
) to up-vote the contribution!