Last updated on October 5th, 2024 at 08:50 pm

Here, We see Count Primes LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Count Primes LeetCode Solution

Problem Statement

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

Example 1:
Input: n = 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.

Example 2:
Input: n = 0
Output: 0

Example 3:
Input: n = 1
Output: 0

1. Count Primes LeetCode Solution C++

class Solution {
public:
	int countPrimes(int n) {
		vector<bool> prime(n + 1, true);
		prime[0] = false;
		prime[1] = false;
		for (int i = 2; i * i <= n; i++) {
			if (prime[i]) {
				for (int j = i * i; j <= n; j += i) {
					prime[j] = false;
				}
			}
		}
		int primeCount = 0;
		for (int i = 2; i < n; i++) {
			if (prime[i]) {
				primeCount++;
			}
		}
		return primeCount;
	}
};

2. Count Primes LeetCode Solution Java

class Solution {
    public int countPrimes(int n) {
        boolean[] seen = new boolean[n];
        int ans = 0;
        for (int num = 2; num < n; num++) {
            if (seen[num]) continue;
            ans += 1;
            for (long mult = (long)num * num; mult < n; mult += num)
                seen[(int)mult] = true;
        }
        return ans;
    }
}

3. Count Primes Solution JavaScript

var countPrimes = function(n) {
    let seen = new Uint8Array(n), ans = 0
    for (let num = 2; num < n; num++) {
        if (seen[num]) continue
        ans++
        for (let mult = num * num; mult < n; mult += num)
            seen[mult] = 1
    }
    return ans
};

4. Count Primes Solution Python

class Solution(object):
    def countPrimes(self, n):
        seen, ans = [0] * n, 0
        for num in range(2, n):
            if seen[num]: continue
            ans += 1
            seen[num*num:n:num] = [1] * ((n - 1) // num - num + 1)
        return ans