# Count Primes LeetCode Solution

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.

# List of all 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

## Count Primes LeetCode SolutionC++

``````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;
}
};```Code language: PHP (php)```

## Count Primes LeetCode SolutionJava

``````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;
}
}```Code language: PHP (php)```

## Count Primes SolutionJavaScript

``````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
};```Code language: JavaScript (javascript)```

## Count Primes SolutionPython

``````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``````
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