Count Primes in Ranges

09 Jun, 2022
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}.
 
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++
Java
Python3
C#
PHP
Javascript
// CPP program to answer queries for count of// primes in given range.#include <bits/stdc++.h>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 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 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# 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.
<?php// PHP 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 = array_fill(0, ($MAX + 1), 0); function buildPrefix(){    global $MAX, $prefix;         // Create a boolean array value in    // prime[i] will "prime[0..n]". A    // finally be false if i is Not a    // prime, else true.    $prime = array_fill(0, ($MAX + 1), true);     for ($p = 2;         $p * $p <= $MAX; $p++)    {         // If prime[p] is not changed,        // then it is a prime        if ($prime[$p] == true)        {             // Update all multiples of p            for ($i = $p * 2;                 $i <= $MAX; $i += $p)                $prime[$i] = false;        }    }     // Build prefix array    // $prefix[0] = $prefix[1] = 0;    for ($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).function query($L, $R){    global $prefix;    return $prefix[$R] -           $prefix[$L - 1];} // Driver codebuildPrefix(); $L = 5;$R = 10;echo query($L, $R) . "\n"; $L = 1;$R = 10;echo query($L, $R) . "\n"; // This code is contributed by mits.?>
<script> // Javascript program to answer queries for// count of primes in given range. let MAX = 10000;   // prefix[i] is going to store count// of primes till i (including i).let prefix = [];   function 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.    let prime = [];    for (let p = 1; p <= MAX +1; p++) {        prime[p] = true;    }       for (let 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 (let i = p * 2; i <= MAX; i += p)        prime[i] = false;    }    }       // Build prefix array    prefix[0] = prefix[1] = 0;    for (let 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).function query(L, R){    return prefix[R] - prefix[L - 1];} // driver program     buildPrefix();    let L = 5, R = 10;    document.write(query(L, R) + "<br/>");       L = 1; R = 10;    document.write(query(L, R)); </script>
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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Mithun Kumar
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array-range-queries
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Prime Number
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