solution(java): 204. Count Primes

204. Count Primes

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;
+    }
+}

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;
+}