Given an array of integers where all the elements are less than 10^6.
The task is to find the difference between the largest and the smallest prime numbers in the array.
Examples:
Input : Array = 1, 2, 3, 5 Output : Difference is 3 Explanation : The largest prime number in the array is 5 and the smallest is 2 So, the difference is 3 Input : Array = 3, 5, 11, 17 Output : Difference is 14
A Simple approach:
In the basic approach, we will check every element of the array whether it is prime or not. Then, select the largest and the smallest prime numbers and print the difference.
Efficient approach: The efficient approach is much similar to the basic approach.
We will try to reduce the time for checking the number against prime by creating a Sieve of Eratosthenes to check whether the number is prime or not in O(1) time.
And then, we will select the largest and the smallest prime numbers and print the difference.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
#define MAX 1000000 bool prime[MAX + 1];
void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
memset (prime, true , sizeof (prime));
// 1 is not prime
prime[1] = false ;
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 ;
}
}
} int findDiff( int arr[], int n)
{ // initial min max value
int min = MAX + 2, max = -1;
for ( int i = 0; i < n; i++) {
// check if the number is prime or not
if (prime[arr[i]] == true ) {
// set the max and min values
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == -1) ? -1 : (max - min);
} // Driver code int main()
{ // create the sieve
SieveOfEratosthenes();
int n = 4;
int arr[n] = { 1, 2, 3, 5 };
int res = findDiff(arr, n);
if (res == -1)
cout << "No prime numbers" << endl;
else
cout << "Difference is " << res << endl;
return 0;
} |
// java implementation of the approach import java.io.*;
class GFG {
static int MAX = 1000000 ;
static boolean prime[] = new boolean [MAX + 1 ];
static void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]" and initialize
// all the entries 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 i= 0 ;i<MAX+ 1 ;i++)
prime[i] = true ;
// 1 is not prime
prime[ 1 ] = false ;
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 ;
}
}
} static int findDiff( int arr[], int n)
{ // initial min max value
int min = MAX + 2 , max = - 1 ;
for ( int i = 0 ; i < n; i++) {
// check if the number is prime or not
if (prime[arr[i]] == true ) {
// set the max and min values
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == - 1 )? - 1 : (max - min);
} // Driver code public static void main (String[] args) {
// create the sieve
SieveOfEratosthenes();
int n = 4 ;
int arr[] = { 1 , 2 , 3 , 5 };
int res = findDiff(arr, n);
if (res == - 1 )
System.out.print( "No prime numbers" ) ;
else
System.out.println( "Difference is " + res);
}
} // This code is contributed by inder_verma.. |
# Python 3 implementation of the approach MAX = 1000000
# Create a boolean array "prime[0..n]" and initialize # all the entries 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():
# 1 is not prime
prime[ 1 ] = False
p = 2
c = 0
while (p * p < = MAX ) :
c + = 1
# If prime[p] is not changed, then it is a prime
if (prime[p] = = True ) :
# Update all multiples of p
for i in range ( p * 2 , MAX + 1 , p):
prime[i] = False
p + = 1
def findDiff(arr, n):
# initial min max value
min = MAX + 2
max = - 1
for i in range (n) :
# check if the number is prime or not
if (prime[arr[i]] = = True ) :
# set the max and min values
#print("arra ",arr[i])
#print("MAX ",max)
#print(" MIN ",min)
if (arr[i] > max ):
max = arr[i]
if (arr[i] < min ):
min = arr[i]
#print(" max ",max)
return - 1 if ( max = = - 1 ) else ( max - min )
# Driver code if __name__ = = "__main__" :
# create the sieve
SieveOfEratosthenes()
n = 4
arr = [ 1 , 2 , 3 , 5 ]
res = findDiff(arr, n)
if (res = = - 1 ):
print ( "No prime numbers" )
else :
print ( "Difference is " ,res )
# this code is contributed by # ChitraNayal |
// C# implementation of the approach using System;
class GFG
{ static int MAX = 1000000;
static bool []prime = new bool [MAX + 1];
static void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]"
// and initialize all the entries 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 i = 0; i < MAX + 1; i++)
prime[i] = true ;
// 1 is not prime
prime[1] = false ;
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 ;
}
}
} static int findDiff( int []arr, int n)
{ // initial min max value
int min = MAX + 2, max = -1;
for ( int i = 0; i < n; i++)
{
// check if the number is prime or not
if (prime[arr[i]] == true )
{
// set the max and min values
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == -1) ? -1 : (max - min);
} // Driver code public static void Main ()
{ // create the sieve
SieveOfEratosthenes();
int n = 4;
int []arr = { 1, 2, 3, 5 };
int res = findDiff(arr, n);
if (res == -1)
Console.WriteLine( "No prime numbers" ) ;
else
Console.WriteLine( "Difference is " + res);
} } // This code is contributed by inder_verma |
<script> // Javascript implementation of above approach MAX = 1000000; prime = new Array(MAX + 1);
function SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
prime.fill( true );
// 1 is not prime
prime[1] = false ;
for ( var 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 ( var i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
} function findDiff(arr, n)
{ // initial min max value
var min = MAX + 2, max = -1;
for ( var i = 0; i < n; i++) {
// check if the number is prime or not
if (prime[arr[i]] == true )
{
// set the max and min values
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == -1)? -1 : (max - min);
} SieveOfEratosthenes(); var n = 4;
var arr = [ 1, 2, 3, 5 ];
var res = findDiff(arr, n);
if (res == -1)
document.write( "No prime numbers" + "<br>" );
else document.write( "Difference is " + res + "<br>" );
// This code is contributed by SoumikMondal </script> |
Difference is 3