Given Q queries where every query consists of an integer D, the task is to find the smallest and the largest prime number with D digits. If no such prime number exists then print -1.
Examples:
Input: Q[] = {2, 5}
Output:
11 97
10007 99991
Input: Q[] = {4, 3, 1}
Output:
1009 9973
101 997
1 7
Approach:
- D digit numbers start from 10(D – 1) and end at 10D – 1.
- Now, the task is to find the smallest and the largest prime number from this range.
- To answer a number of queries for prime numbers, Sieve of Eratosthenes can be used to answer whether a number is prime or not.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAX 100000 bool prime[MAX + 1]; void SieveOfEratosthenes() { // Create a boolean array "prime[0..n]" and initialize // all entries it as true. A value in prime[i] will // finally be false if i is Not a prime, else true. 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 greater than or // equal to the square of it // numbers which are multiple of p and are // less than p^2 are already been marked. for ( int i = p * p; i <= MAX; i += p) prime[i] = false ; } } } // Function to return the smallest prime // number with d digits int smallestPrime( int d) { int l = pow (10, d - 1); int r = pow (10, d) - 1; for ( int i = l; i <= r; i++) { // check if prime if (prime[i]) { return i; } } return -1; } // Function to return the largest prime // number with d digits int largestPrime( int d) { int l = pow (10, d - 1); int r = pow (10, d) - 1; for ( int i = r; i >= l; i--) { // check if prime if (prime[i]) { return i; } } return -1; } // Driver code int main() { SieveOfEratosthenes(); int queries[] = { 2, 5 }; int q = sizeof (queries) / sizeof (queries[0]); // Perform queries for ( int i = 0; i < q; i++) { cout << smallestPrime(queries[i]) << " " << largestPrime(queries[i]) << endl; } return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { static int MAX = 100000 ; static boolean []prime = new boolean [MAX + 1 ]; static void SieveOfEratosthenes() { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. // A value in prime[i] will finally be false // if i is Not a prime, else true. for ( int i = 0 ; i < MAX + 1 ; i++) { prime[i] = 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 greater than or // equal to the square of it // numbers which are multiple of p and are // less than p^2 are already been marked. for ( int i = p * p; i <= MAX; i += p) prime[i] = false ; } } } // Function to return the smallest prime // number with d digits static int smallestPrime( int d) { int l = ( int ) Math.pow( 10 , d - 1 ); int r = ( int ) Math.pow( 10 , d) - 1 ; for ( int i = l; i <= r; i++) { // check if prime if (prime[i]) { return i; } } return - 1 ; } // Function to return the largest prime // number with d digits static int largestPrime( int d) { int l = ( int ) Math.pow( 10 , d - 1 ); int r = ( int ) Math.pow( 10 , d) - 1 ; for ( int i = r; i >= l; i--) { // check if prime if (prime[i]) { return i; } } return - 1 ; } // Driver code public static void main(String[] args) { SieveOfEratosthenes(); int queries[] = { 2 , 5 }; int q = queries.length; // Perform queries for ( int i = 0 ; i < q; i++) { System.out.println(smallestPrime(queries[i]) + " " + largestPrime(queries[i])); } } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach from math import sqrt MAX = 100000 # Create a boolean array "prime[0..n]" and # initialize all entries it as true. # A value in prime[i] will finally be false # if i is Not a prime, else true. prime = [ True ] * ( MAX + 1 ) def SieveOfEratosthenes() : for p in range ( 2 , int (sqrt( MAX )) + 1 ) : # If prime[p] is not changed, # then it is a prime if (prime[p] = = True ) : # Update all multiples of p greater than or # equal to the square of it # numbers which are multiple of p and are # less than p^2 are already been marked. for i in range (p * p, MAX + 1 , p) : prime[i] = False ; # Function to return the smallest prime # number with d digits def smallestPrime(d) : l = 10 * * (d - 1 ); r = ( 10 * * d) - 1 ; for i in range (l, r + 1 ) : # check if prime if (prime[i]) : return i; return - 1 ; # Function to return the largest prime # number with d digits def largestPrime(d) : l = 10 * * (d - 1 ); r = ( 10 * * d) - 1 ; for i in range (r, l , - 1 ) : # check if prime if (prime[i]) : return i; return - 1 ; # Driver code if __name__ = = "__main__" : SieveOfEratosthenes(); queries = [ 2 , 5 ]; q = len (queries); # Perform queries for i in range (q) : print (smallestPrime(queries[i]), " " , largestPrime(queries[i])); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System; class GFG { static int MAX = 100000; static bool []prime = new bool [MAX + 1]; static void SieveOfEratosthenes() { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. // A value in prime[i] will finally be false // if i is Not a prime, else true. for ( int i = 0; i < MAX + 1; i++) { prime[i] = 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 greater than // or equal to the square of it // numbers which are multiple of p and are // less than p^2 are already been marked. for ( int i = p * p; i <= MAX; i += p) prime[i] = false ; } } } // Function to return the smallest prime // number with d digits static int smallestPrime( int d) { int l = ( int ) Math.Pow(10, d - 1); int r = ( int ) Math.Pow(10, d) - 1; for ( int i = l; i <= r; i++) { // check if prime if (prime[i]) { return i; } } return -1; } // Function to return the largest prime // number with d digits static int largestPrime( int d) { int l = ( int ) Math.Pow(10, d - 1); int r = ( int ) Math.Pow(10, d) - 1; for ( int i = r; i >= l; i--) { // check if prime if (prime[i]) { return i; } } return -1; } // Driver code public static void Main(String[] args) { SieveOfEratosthenes(); int []queries = { 2, 5 }; int q = queries.Length; // Perform queries for ( int i = 0; i < q; i++) { Console.WriteLine(smallestPrime(queries[i]) + " " + largestPrime(queries[i])); } } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach const MAX = 100000; // Create a boolean array // "prime[0..n]" and initialize // all entries it as true. // A value in prime[i] will // finally be false if i is Not a prime, // else true. let prime = new Array(MAX + 1).fill( true ); function SieveOfEratosthenes() { 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 // greater than or // equal to the square of it // numbers which are multiple // of p and are // less than p^2 are already // been marked. for (let i = p * p; i <= MAX; i += p) prime[i] = false ; } } } // Function to return the smallest prime // number with d digits function smallestPrime(d) { let l = Math.pow(10, d - 1); let r = Math.pow(10, d) - 1; for (let i = l; i <= r; i++) { // check if prime if (prime[i]) { return i; } } return -1; } // Function to return the largest prime // number with d digits function largestPrime(d) { let l = Math.pow(10, d - 1); let r = Math.pow(10, d) - 1; for (let i = r; i >= l; i--) { // check if prime if (prime[i]) { return i; } } return -1; } // Driver code SieveOfEratosthenes(); let queries = [ 2, 5 ]; let q = queries.length; // Perform queries for (let i = 0; i < q; i++) { document.write(smallestPrime(queries[i]) + " " + largestPrime(queries[i]) + "<br>" ); } </script> |
Output:
11 97 10007 99991
Time Complexity: O(MAX*log(log(MAX)) + q*(r – l))
Auxiliary Space: O(MAX)
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