Sieve of Eratosthenes Find All Primes in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Sieve for n=10

What is the output array of primes when running the Sieve of Eratosthenes for n=10?

DSA Javascript

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

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

B [2, 3, 4, 5, 7]

C [2, 3, 5, 7]

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

💡 Hint

Remember that 1 is not a prime number and multiples of primes are marked false.

The sieve marks non-primes false. For n=10, primes are 2, 3, 5, and 7 only.

🧠 Conceptual

intermediate

Why start crossing out multiples from i*i?

In the Sieve of Eratosthenes, why do we start marking multiples of a prime i from i*i instead of 2*i?

A Because smaller multiples of i are already marked by smaller primes.

B Because starting from 2*i would miss some prime numbers.

C Because 2*i might not be a multiple of i.

D Because i*i is the smallest prime number.

💡 Hint

Think about what happens to multiples less than i*i.

Multiples less than i*i are already marked by smaller primes, so starting at i*i avoids redundant work.

🔧 Debug

advanced

Identify the error in this sieve implementation

What error will this code produce when run for n=10?

DSA Javascript

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

A No error, returns correct primes

B Returns incorrect primes missing 7

C Returns primes including 1

D TypeError: Cannot set property '10' of undefined

💡 Hint

Check the size of the primes array and the indices accessed.

The array size is n, but code accesses primes[n], which is undefined, causing a TypeError.

❓ Predict Output

advanced

Output of sieve for n=20 with a bug

What is the output of this buggy sieve code for n=20?

DSA Javascript

function sieve(n) {
  const primes = Array(n + 1).fill(true);
  primes[0] = false;
  primes[1] = false;
  for (let i = 2; i <= n; i++) {
    if (primes[i]) {
      for (let j = i * 2; j <= n; j += i) {
        primes[j] = false;
      }
    }
  }
  return primes.map((isPrime, index) => isPrime ? index : null).filter(x => x !== null);
}
console.log(sieve(20));

A [2, 3, 5, 7, 9, 11, 13, 15, 17, 19]

B [2, 3, 5, 7, 9, 11, 13, 17, 19]

C [2, 3, 5, 7, 11, 13, 17, 19, 20]

D [2, 3, 5, 7, 11, 13, 17, 19]

💡 Hint

Check where the inner loop starts crossing out multiples.

Starting from i*2 instead of i*i causes some composites like 9 and 15 to remain true incorrectly.

🚀 Application

expert

Count primes in a large range efficiently

You want to count how many prime numbers are between 1 and 1,000,000 using the Sieve of Eratosthenes. Which approach is best to optimize memory and speed?

A Use a list to store all numbers and remove multiples of each prime found, starting from 2.

B Use a boolean array of size 1,000,001 and mark multiples starting from i*i, skipping even numbers after 2.

C Check each number from 2 to 1,000,000 individually for primality by dividing by all smaller numbers.

D Use recursion to mark multiples of primes up to 1,000,000.

💡 Hint

Think about memory usage and avoiding unnecessary checks.

Using a boolean array and starting from i*i is the classic efficient sieve. Skipping even numbers after 2 reduces work and memory.

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