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# Count Primes in Ranges

Given a range [L, R], we need to find the count of total numbers of prime numbers in the range [L, R] where 0 <= L <= R < 10000. Consider that there are a large number of queries for different ranges.
Examples:

Input : Query 1 : L = 1, R = 10
Query 2 : L = 5, R = 10
Output : 4
2
Explanation
Primes in the range L = 1 to R = 10 are
{2, 3, 5, 7}. Therefore for query, answer
is 4 {2, 3, 5, 7}.
For the second query, answer is 2 {5, 7}.

Recommended Practice

A simple solution is to do the following for every query [L, R]. Traverse from L to R, check if current number is prime. If yes, increment the count. Finally, return the count.
An efficient solution is to use Sieve of Eratosthenes to find all primes up to the given limit. Then we compute a prefix array to store counts till every value before limit. Once we have a prefix array, we can answer queries in O(1) time. We just need to return prefix[R] – prefix[L-1].

## C++

 // CPP program to answer queries for count of// primes in given range.#include using namespace std; const int MAX = 10000; // prefix[i] is going to store count of primes// till i (including i).int prefix[MAX + 1]; void buildPrefix(){    // Create a boolean array "prime[0..n]". A    // value in prime[i] will finally be false    // if i is Not a prime, else true.    bool prime[MAX + 1];    memset(prime, true, sizeof(prime));     for (int p = 2; p * p <= MAX; p++) {         // If prime[p] is not changed, then        // it is a prime        if (prime[p] == true) {             // Update all multiples of p            for (int i = p * 2; i <= MAX; i += p)                prime[i] = false;        }    }     // Build prefix array    prefix[0] = prefix[1] = 0;    for (int p = 2; p <= MAX; p++) {        prefix[p] = prefix[p - 1];        if (prime[p])            prefix[p]++;    }} // Returns count of primes in range from L to// R (both inclusive).int query(int L, int R){    return prefix[R] - prefix[L - 1];} // Driver codeint main(){    buildPrefix();     int L = 5, R = 10;    cout << query(L, R) << endl;     L = 1, R = 10;    cout << query(L, R) << endl;     return 0;}

## Java

 // Java program to answer queries for// count of primes in given range.import java.util.*; class GFG {     static final int MAX = 10000; // prefix[i] is going to store count// of primes till i (including i).static int prefix[] = new int[MAX + 1]; static void buildPrefix() {         // Create a boolean array "prime[0..n]". A    // value in prime[i] will finally be false    // if i is Not a prime, else true.    boolean prime[] = new boolean[MAX + 1];    Arrays.fill(prime, true);     for (int p = 2; p * p <= MAX; p++) {     // If prime[p] is not changed, then    // it is a prime    if (prime[p] == true) {         // Update all multiples of p        for (int i = p * 2; i <= MAX; i += p)        prime[i] = false;    }    }     // Build prefix array    prefix[0] = prefix[1] = 0;    for (int p = 2; p <= MAX; p++) {    prefix[p] = prefix[p - 1];    if (prime[p])        prefix[p]++;    }} // Returns count of primes in range// from L to R (both inclusive).static int query(int L, int R){    return prefix[R] - prefix[L - 1];} // Driver codepublic static void main(String[] args) {         buildPrefix();    int L = 5, R = 10;    System.out.println(query(L, R));     L = 1; R = 10;    System.out.println(query(L, R));}} // This code is contributed by Anant Agarwal.

## Python3

 # Python3 program to answer queries for# count of primes in given range.MAX = 10000 # prefix[i] is going to# store count of primes# till i (including i).prefix =[0]*(MAX + 1) def buildPrefix():         # Create a boolean array value in    # prime[i] will "prime[0..n]". A    # finally be false if i is Not a    # prime, else true.    prime = [1]*(MAX + 1)     p = 2    while(p * p <= MAX):         # If prime[p] is not changed,        # then it is a prime        if (prime[p] == 1):             # Update all multiples of p            i = p * 2            while(i <= MAX):                prime[i] = 0                i += p        p+=1     # Build prefix array    # prefix[0] = prefix[1] = 0;    for p in range(2,MAX+1):        prefix[p] = prefix[p - 1]        if (prime[p]==1):            prefix[p]+=1 # Returns count of primes# in range from L to# R (both inclusive).def query(L, R):    return prefix[R]-prefix[L - 1] # Driver codeif __name__=='__main__':    buildPrefix()     L = 5    R = 10    print(query(L, R))     L = 1    R = 10    print(query(L, R)) # This code is contributed by mits.

## C#

 // C# program to answer// queries for count of// primes in given range.using System; class GFG{static int MAX = 10000; // prefix[i] is going// to store count of// primes till i (including i).static int[] prefix = new int[MAX + 1]; static void buildPrefix(){         // Create a boolean array    // "prime[0..n]". A value    // in prime[i] will finally    // be false if i is Not a    // prime, else true.    bool[] prime = new bool[MAX + 1];     for (int p = 2;             p * p <= MAX; p++)    {     // If prime[p] is    // not changed, then    // it is a prime    if (prime[p] == false)    {         // Update all        // multiples of p        for (int i = p * 2;                 i <= MAX; i += p)        prime[i] = true;    }    }     // Build prefix array    prefix[0] = prefix[1] = 0;    for (int p = 2; p <= MAX; p++)    {        prefix[p] = prefix[p - 1];        if (prime[p] == false)            prefix[p]++;    }} // Returns count of primes// in range from L to R// (both inclusive).static int query(int L, int R){    return prefix[R] -           prefix[L - 1];} // Driver codepublic static void Main(){    buildPrefix();    int L = 5, R = 10;    Console.WriteLine(query(L, R));     L = 1; R = 10;    Console.WriteLine(query(L, R));}} // This code is contributed// by mits.



## Javascript



Output:

2
4

Time Complexity: O(n*log(log(n)))

Auxiliary Space: O(n)

Here, n is the size of the prime array, Which is MAX here

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