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This Java code provides a solution for LeetCode problem godkingjay#204,…
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… which requires counting the number of prime numbers strictly less than a given integer n. The code employs the efficient Sieve of Eratosthenes algorithm to identify and count prime numbers within the specified range. It's a practical and optimized solution for counting prime numbers and can be easily integrated into other Java projects.

The code is well-documented and includes sample test cases for verification. It offers a clear and concise implementation of the Sieve of Eratosthenes approach, making it a valuable resource for solving similar problems involving prime numbers in various programming scenarios.
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Jaiyadav88 committed Oct 11, 2023
1 parent 18fedc0 commit 68d9529
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37 changes: 37 additions & 0 deletions Medium/204.CountPrimes/Count-Primes.java
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/*
* LeetCode 204.
* Problem Statement-Given an integer n, return the number of prime numbers that are strictly less than n
*
* Test Cases:
*
* 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
*/

//Solution of Sieve of Eratosthenes approach
class Solution {
public int countPrimes(int n) {
//using seive of eratosthenes
boolean prime[]=new boolean[n+1];int c=0;
Arrays.fill(prime,true);
prime[0]=false;
for(int i=2;i*i<=n;i++)
{
if(prime[i]==true)
{
for(int j=i*i;j<=n;j=j+i)
prime[j]=false;
}
}
for(int i=2;i<n;i++)
if(prime[i])
c++;
return c;
}
}
38 changes: 38 additions & 0 deletions Medium/204.CountPrimes/README.md
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# LeetCode 204 - Count Primes

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

### Test Cases

#### 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`

## Solution
This solution utilizes the Sieve of Eratosthenes algorithm to efficiently count the number of prime numbers less than `n`.

### Pseudocode
```java
public int countPrimes(int n) {
// Using Sieve of Eratosthenes
boolean prime[] = new boolean[n + 1];
int count = 0;
Arrays.fill(prime, true);
prime[0] = false;
for (int i = 2; i * i <= n; i++) {
if (prime[i] == true) {
for (int j = i * i; j <= n; j += i)
prime[j] = false;
}
}
for (int i = 2; i < n; i++)
if (prime[i])
count++;
return count;
}

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