How do you solve Count Primes on LeetCode?

Count Primes (LeetCode #204) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log log n) time complexity and O(n) space complexity. Key insight: Understanding prime numbers is crucial for optimizing algorithms.

What is the brute force solution for Count Primes?

The brute force approach for Count Primes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each number less than n to see if it is prime. This is straightforward but inefficient, as it involves checking each number individually. The optimal Optimal Solution approach improves this to O(n log log n).

What algorithm or data structure is used to solve Count Primes?

Count Primes uses the following concepts: Array, Math, Enumeration, Number Theory. The recommended patterns to study are: Sieve of Eratosthenes, Array.

What are the interview tips for Count Primes?

Always start with a brute force solution to demonstrate your understanding. Explain your thought process clearly, especially when transitioning to an optimized approach. Practice implementing the Sieve of Eratosthenes, as it's a common algorithm in interviews.

What are common mistakes when solving Count Primes?

Not recognizing that 0 and 1 are not prime numbers. Failing to optimize checks by using the Sieve of Eratosthenes instead of brute force.

#204

Count Primes

Medium

ArrayMathEnumerationNumber TheorySieve of EratosthenesArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log log n)
Space	O(1)	O(n)

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Intuition

Time O(n log log n)Space O(n)

The Sieve of Eratosthenes is an efficient algorithm to find all primes up to a given limit. It works by iteratively marking the multiples of each prime number starting from 2.

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Algorithm

4 steps

1Step 1: Create a boolean array 'is_prime' of size n initialized to True.
2Step 2: Set is_prime[0] and is_prime[1] to False (0 and 1 are not primes).
3Step 3: Loop from 2 to the square root of n. For each number, if it is still marked as prime, mark all its multiples as not prime.
4Step 4: Count the number of True values in 'is_prime' array, which represents the prime numbers.

solution.py12 lines

1# Full working Python code
2
3def countPrimes(n):
4    if n < 3:
5        return 0
6    is_prime = [True] * n
7    is_prime[0], is_prime[1] = False, False
8    for i in range(2, int(n**0.5) + 1):
9        if is_prime[i]:
10            for j in range(i * i, n, i):
11                is_prime[j] = False
12    return sum(is_prime)

ℹ

Complexity note: The time complexity is O(n log log n) due to the Sieve of Eratosthenes marking multiples, which is much faster than the brute force method. The space complexity is O(n) for storing the boolean array.

1Understanding prime numbers is crucial for optimizing algorithms.
2The Sieve of Eratosthenes is a classic example of an efficient algorithm that reduces unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.