Sieve of Eratosthenes Find All Primes in DSA Python - 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.

Specifically, how does the work change when we increase the limit up to which we find primes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

    def sieve_of_eratosthenes(n):
        primes = [True] * (n + 1)
        primes[0], primes[1] = False, False
        p = 2
        while p * p <= n:
            if primes[p]:
                for i in range(p * p, n + 1, p):
                    primes[i] = False
            p += 1
        return [x for x in range(n + 1) if primes[x]]
    

This code finds all prime numbers up to n by marking multiples of each prime as not prime.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Marking multiples of each prime number as not prime inside the inner for-loop.
• How many times: The outer while-loop runs roughly up to the square root of n, and the inner loop marks multiples starting from p squared up to n.

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 to 30 operations
100	About 300 to 400 operations
1000	About 4000 to 5000 operations

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

Final Time Complexity

Time Complexity: O(n log log n)

This means the time needed 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 algorithm only marks multiples of primes, skipping many numbers quickly, so it does much less work than checking all pairs.

Interview Connect

Understanding this algorithm's time complexity shows you can analyze efficient number-based algorithms, a useful skill for many coding challenges.

Self-Check

"What if we started marking multiples from 2*p instead of p*p? 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 a list of numbers in ascending order

C. To find the greatest common divisor of two numbers

D. To calculate the factorial of a number

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, leaving primes unmarked up to the 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: Sieve marks multiples to find primes fast [OK]

Common Mistakes:

• Confusing it with sorting algorithms
• Thinking it calculates factorials
• Assuming it finds divisors instead of primes

2. Which of the following Python code snippets correctly initializes the sieve list for finding primes up to n?

easy

A. sieve = [False] * n

B. sieve = list(range(n))

C. sieve = [True] * (n+1)

D. sieve = [0] * (n-1)

Solution

1. Step 1: Understand sieve initialization

   The sieve list should mark all numbers as potentially prime initially, so all True.

2. Step 2: Check correct length

   It must include index n, so length is n+1.

3. Final Answer:

   sieve = [True] * (n+1) -> Option C

4. Quick Check:

   Initialize all True for 0 to n [OK]

Hint: Use True for all numbers including n [OK]

Common Mistakes:

• Using False instead of True initially
• Not including n in the list length
• Using range instead of boolean list

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

n = 10
sieve = [True] * (n+1)
sieve[0], sieve[1] = False, False
for i in range(2, int(n**0.5)+1):
    if sieve[i]:
        for j in range(i*i, n+1, i):
            sieve[j] = False
primes = [i for i in range(n+1) if sieve[i]]
print(primes)

medium

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

B. [2, 3, 5, 7]

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

D. [0, 2, 3, 5, 7]

Solution

1. Step 1: Initialize sieve and mark 0,1 as non-prime

   Indices 0 and 1 are set to False, others True initially.

2. Step 2: Mark multiples of primes as False

   For i=2 and i=3 (since sqrt(10) ~3), multiples 4,6,8,9,10 are marked False.

3. Step 3: Collect primes

   Numbers with True are 2,3,5,7.

4. Final Answer:

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

5. Quick Check:

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

Hint: Remember to exclude 0 and 1 as non-prime [OK]

Common Mistakes:

• Including 1 as prime
• Forgetting to start inner loop at i*i
• Marking wrong multiples

4. Identify the error in this Sieve of Eratosthenes code snippet:

n = 15
sieve = [True] * (n+1)
sieve[0], sieve[1] = False, False
for i in range(2, n):
    if sieve[i]:
        for j in range(i*2, n+1, i):
            sieve[j] = False
primes = [i for i in range(n+1) if sieve[i]]
print(primes)

medium

A. Inner loop should start from i*i, not i*2

B. sieve list length should be n, not n+1

C. sieve[0] and sieve[1] should be True

D. Outer loop should run until n+1, not n

Solution

1. Step 1: Check inner loop start index

   Starting from i*2 marks some multiples multiple times, inefficient.

2. Step 2: Correct start is i*i

   Because smaller multiples are already marked by smaller primes.

3. Final Answer:

   Inner loop should start from i*i, not i*2 -> Option A

4. Quick Check:

   Start inner loop at i*i for efficiency [OK]

Hint: Start marking multiples from i squared [OK]

Common Mistakes:

• Starting inner loop at i*2 wastes time
• Incorrect sieve length causes index errors
• Marking 0 or 1 as prime

5. Using the Sieve of Eratosthenes, how many prime numbers are there between 1 and 30 (inclusive)?

hard

A. 8

B. 11

C. 12

D. 10

Solution

1. Step 1: List primes up to 30 using sieve logic

   Primes are 2,3,5,7,11,13,17,19,23,29.

2. Step 2: Count the primes

   Count is 10 primes between 1 and 30 inclusive.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Count primes ≤30 = 10 [OK]

Hint: Count primes carefully, exclude 1 and non-primes [OK]

Common Mistakes:

• Including 1 as prime
• Forgetting to include 29
• Miscounting primes between 20 and 30