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In this post, you will find the solution for the **Count Primes** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Count Primes– LeetCode Problem

Count Primes– LeetCode Problem

**Problem:**

Given an integer `n`

, return *the number of prime numbers that are strictly less than* `n`

.

**Example 1:**

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

**Example 2:**

Input:n = 0Output:0

**Example 3:**

Input:n = 1Output:0

**Constraints:**

`0 <= n <= 5 * 10`

^{6}

Count Primes– LeetCode Solutions

Count PrimesSolution in C++:

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

Count PrimesSolution in Java:

class Solution { public int countPrimes(int n) { if (n <= 2) return 0; int ans = 0; boolean[] prime = new boolean[n]; Arrays.fill(prime, 2, n, true); for (int i = 0; i < Math.sqrt(n); ++i) if (prime[i]) for (int j = i * i; j < n; j += i) prime[j] = false; for (final boolean p : prime) if (p) ++ans; return ans; } }

Count PrimesSolution in Python:

class Solution: def countPrimes(self, n: int) -> int: if n <= 2: return 0 isPrime = [False] * 2 + [True] * (n - 2) for i in range(2, int(n**0.5) + 1): if isPrime[i]: for j in range(i * i, n, i): isPrime[j] = False return sum(isPrime)

**Time:***O*(*n*loglog*n*)**Space:**O(*n*)