Count Primes

Count Primes - Problem

Given an integer n, return the number of prime numbers that are strictly less than n.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Input & Output

Example 1 — Small Range

$ Input: n = 10

› Output: 4

💡 Note: Primes less than 10 are [2, 3, 5, 7], so count = 4

Example 2 — Very Small Range

$ Input: n = 2

› Output: 0

💡 Note: No primes less than 2, so count = 0

Example 3 — Edge Case

$ Input: n = 3

› Output: 1

💡 Note: Only prime less than 3 is [2], so count = 1

Constraints

0 ≤ n ≤ 5 × 10⁶

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15 f Facebook 12

The Sieve of Eratosthenes is the optimal approach. Key insight: instead of testing each number for primality, mark all multiples of each prime as composite. Best approach uses optimized sieve with Time: O(n log log n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each number from 2 to n-1, check if it's prime by testing divisibility from 2 to sqrt(num). Count all numbers that pass the primality test.

Optimized Sieve

⏱️ Time: O(n log log n) Space: O(n)

Optimize the sieve by only checking odd numbers after 2, starting marking from i² instead of 2i, and using other mathematical properties to reduce operations.

Sieve of Eratosthenes

⏱️ Time: O(n log log n) Space: O(n)

Create a boolean array and mark all multiples of each prime as composite. The remaining unmarked numbers are primes. This avoids redundant primality testing.

Brute Force — Algorithm Steps

Iterate through numbers 2 to n-1
For each number, check if it has any divisors from 2 to sqrt(num)
Count numbers that have no divisors (primes)

Visualization

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Step-by-Step Walkthrough

Test 2

No divisors from 2 to √2, so 2 is prime

Test 3

No divisors from 2 to √3, so 3 is prime

Test 4

4 ÷ 2 = 2, so 4 is not prime

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    if (n <= 2) return 0;
    
    int count = 0;
    for (int num = 2; num < n; num++) {
        int isPrime = 1;
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                isPrime = 0;
                break;
            }
        }
        if (isPrime) count++;
    }
    
    return count;
}

int main() {
    int n;
    scanf("%d", &n);
    printf("%d\n", solution(n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n numbers, we check up to sqrt(n) divisors

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

125.0K Views

Medium Frequency

~15 min Avg. Time

5.8K Likes

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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