Sieve of Eratosthenes Find All Primes in DSA Javascript - Time & Space Complexity

Time Complexity: Sieve of Eratosthenes Find All Primes

O(n log log n)

Understanding Time Complexity

We want to understand how the time needed to find all prime numbers grows as the input number gets bigger.

How does the work increase when we check more numbers?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function sieveOfEratosthenes(n) {
  const primes = new 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;
}
    

This code marks non-prime numbers as false to find all primes up to n.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop runs from 2 to √n, and the inner loop marks multiples of i as false.
• How many times: Outer loop runs about √n times; inner loop runs fewer times as i grows, marking multiples.

How Execution Grows With Input

As n grows, the number of operations grows a bit faster than n but much slower than n squared.

Input Size (n)	Approx. Operations
10	About 20-30 operations
100	About 500-600 operations
1000	About 8,000-10,000 operations

Pattern observation: The operations grow roughly proportional to n times the logarithm of logarithm of n, which is slower than n log n but faster than n.

Final Time Complexity

Time Complexity: O(n log log n)

This means the time grows a bit faster than n but much slower than n squared, making it efficient for large n.

Common Mistake

[X] Wrong: "The algorithm checks every number against all smaller numbers, so it must be O(n²)."

[OK] Correct: The inner loop skips many numbers by jumping in steps of i, and the outer loop only goes up to √n, so it does much less work than n squared.

Interview Connect

Understanding this time complexity shows you can analyze nested loops carefully and recognize efficient number checking methods, a useful skill in many coding challenges.

Self-Check

"What if we started the inner loop from 2*i instead of i*i? How would the time complexity change?"

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