Sieve of Eratosthenes Find All Primes in DSA Javascript - Build from Scratch

Sieve of Eratosthenes Find All Primes

📖 Scenario: You are helping a teacher prepare a list of prime numbers for a math class. Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. The teacher wants to find all prime numbers up to a certain number using a fast method called the Sieve of Eratosthenes.

🎯 Goal: Build a program that finds all prime numbers up to a given number n using the Sieve of Eratosthenes method and prints them as a list.

📋 What You'll Learn

Create an array called isPrime to track if numbers are prime

Create a variable called n and set it to 30

Use a loop to mark non-prime numbers in isPrime

Collect all prime numbers into an array called primes

Print the primes array

💡 Why This Matters

🌍 Real World

Finding prime numbers is useful in cryptography, computer security, and number theory research.

💼 Career

Understanding the Sieve of Eratosthenes helps in algorithm design and optimization tasks in software development.

1

Create the initial array to track primes

Create a variable called n and set it to 30. Then create an array called isPrime of length n + 1 and fill it with true values.

DSA Javascript

const n = 30;
// Your code here

Hint

Use new Array(n + 1).fill(true) to create the array.

2

Set 0 and 1 as not prime

Set isPrime[0] and isPrime[1] to false because 0 and 1 are not prime numbers.

DSA Javascript

const n = 30;
const isPrime = new Array(n + 1).fill(true);
// Your code here

Hint

Remember 0 and 1 are not prime numbers.

3

Mark multiples of each prime as not prime

Use a for loop with variable i starting from 2 up to Math.sqrt(n). Inside it, use another for loop with variable j starting from i * i up to n, incrementing by i. Set isPrime[j] to false inside the inner loop if isPrime[i] is true.

DSA Javascript

const n = 30;
const isPrime = new Array(n + 1).fill(true);
isPrime[0] = false;
isPrime[1] = false;
// Your code here

Hint

Use nested loops to mark multiples of primes as not prime.

4

Collect and print all prime numbers

Create an empty array called primes. Use a for loop with variable k from 2 to n. Inside the loop, if isPrime[k] is true, push k into primes. Finally, print the primes array using console.log(primes).

DSA Javascript

const n = 30;
const isPrime = new Array(n + 1).fill(true);
isPrime[0] = false;
isPrime[1] = false;
for (let i = 2; i <= Math.sqrt(n); i++) {
    if (isPrime[i]) {
        for (let j = i * i; j <= n; j += i) {
            isPrime[j] = false;
        }
    }
}
// Your code here

Hint

Collect all numbers marked true in isPrime and print them.

Practice

1. What is the main purpose of the Sieve of Eratosthenes algorithm?

easy

A. To find all prime numbers up to a given limit

B. To sort an array of numbers

C. To find the greatest common divisor of two numbers

D. To reverse a string

Solution

1. Step 1: Understand the algorithm's goal

   The Sieve of Eratosthenes is designed to identify prime numbers efficiently.

2. Step 2: Recognize the output

   It marks multiples of primes as non-prime to find all primes up to a limit.

3. Final Answer:

   To find all prime numbers up to a given limit -> Option A

4. Quick Check:

   Sieve of Eratosthenes = Find primes [OK]

Hint: Remember: sieve filters out non-primes to find primes [OK]

Common Mistakes:

• Confusing it with sorting algorithms
• Thinking it finds factors instead of primes
• Assuming it works on strings

2. Which of the following JavaScript code snippets correctly initializes the sieve array for numbers up to n?

easy

A. const sieve = Array(n).fill(true); sieve[0] = sieve[1] = true;

B. const sieve = new Array(n).fill(false); sieve[0] = sieve[1] = true;

C. const sieve = new Array(n + 1).fill(true); sieve[0] = sieve[1] = false;

D. const sieve = new Array(n + 1).fill(false); sieve[0] = sieve[1] = false;

Solution

1. Step 1: Understand sieve initialization

   The sieve array should mark all numbers as true initially, except 0 and 1 which are not prime.

2. Step 2: Check each option

   const sieve = new Array(n + 1).fill(true); sieve[0] = sieve[1] = false; correctly creates an array of size n+1 filled with true, then sets 0 and 1 to false.

3. Final Answer:

   const sieve = new Array(n + 1).fill(true); sieve[0] = sieve[1] = false; -> Option C

4. Quick Check:

   Initialize sieve with true, except 0 and 1 false [OK]

Hint: Sieve size is n+1; 0 and 1 are not prime [OK]

Common Mistakes:

• Using size n instead of n+1
• Filling with false instead of true
• Marking 0 and 1 as true

3. What will be the output of the following code snippet?

function sieve(n) {
  const sieve = new Array(n + 1).fill(true);
  sieve[0] = sieve[1] = false;
  for (let i = 2; i * i <= n; i++) {
    if (sieve[i]) {
      for (let j = i * i; j <= n; j += i) {
        sieve[j] = false;
      }
    }
  }
  return sieve.map((val, idx) => val ? idx : null).filter(x => x);
}
console.log(sieve(10));

medium

A. [1, 2, 3, 5, 7]

B. [2, 3, 5, 7]

C. [2, 3, 4, 5, 7]

D. [2, 3, 5, 7, 9]

Solution

1. Step 1: Understand sieve marking

   The code marks multiples of primes as false, starting from i*i, up to n=10.

2. Step 2: Identify primes up to 10

   Primes are 2, 3, 5, 7. Numbers 1, 4, 6, 8, 9, 10 are marked false.

3. Final Answer:

   [2, 3, 5, 7] -> Option B

4. Quick Check:

   Primes up to 10 = [2, 3, 5, 7] [OK]

Hint: Primes under 10 are 2,3,5,7 only [OK]

Common Mistakes:

• Including 1 as prime
• Including non-primes like 4 or 9
• Off-by-one errors in loops

4. Identify the error in the following Sieve of Eratosthenes implementation:

function sieve(n) {
  const sieve = new Array(n).fill(true);
  sieve[0] = sieve[1] = false;
  for (let i = 2; i <= n; i++) {
    if (sieve[i]) {
      for (let j = i * 2; j < n; j += i) {
        sieve[j] = false;
      }
    }
  }
  return sieve.map((val, idx) => val ? idx : null).filter(x => x);
}

medium

A. Array size should be n+1 to include number n

B. Inner loop should start from i*3 instead of i*2

C. Outer loop should run only up to sqrt(n)

D. sieve[0] and sieve[1] should be true

Solution

1. Step 1: Check array size

   The array is created with size n, so index n is out of range and not included.

2. Step 2: Understand loop bounds

   To include number n, array size must be n+1; otherwise, sieve[n] is undefined.

3. Final Answer:

   Array size should be n+1 to include number n -> Option A

4. Quick Check:

   Array size = n+1 to cover all numbers [OK]

Hint: Array size must cover all indices up to n [OK]

Common Mistakes:

• Using array size n instead of n+1
• Looping outer loop beyond sqrt(n)
• Starting inner loop from i*3 incorrectly

5. You want to find all prime numbers between 50 and 100 using the Sieve of Eratosthenes. Which approach is correct?

hard

A. Run sieve up to 150 and ignore primes below 50

B. Run sieve only from 50 to 100 directly

C. Run sieve up to 50, then add primes from 51 to 100 manually

D. Run sieve up to 100, then filter primes >= 50 from the result

Solution

1. Step 1: Understand sieve range requirement

   The sieve finds primes from 2 up to the limit; partial ranges are not supported directly.

2. Step 2: Filter primes after full sieve

   Run sieve up to 100, then select primes >= 50 from the result for correct output.

3. Final Answer:

   Run sieve up to 100, then filter primes >= 50 from the result -> Option D

4. Quick Check:

   Sieve full range, then filter needed primes [OK]

Hint: Sieve full range, then filter start point [OK]

Common Mistakes:

• Trying to sieve partial ranges directly
• Ignoring primes below 50 without full sieve
• Manually adding primes without verification