Find the position of the given Prime Number
Last Updated :
21 Nov, 2021
Given a number N which is a prime number, the task is to find the position of the given prime number in the series of Prime Numbers.
Examples :
Input: N = 11
Output: 5
Explanation:
The prime numbers are 2, 3, 5, 7, 11, 13, 17, ….
Therefore, the position of 11 in this series is 5.
Input: N = 13
Output: 6
Naive Approach: The naive approach for this problem is for the given input, compute the prime numbers which are less than that number and keep a track of the number of primes less than the given N. If the count is K, then K + 1 would be the answer. The time complexity for this approach is quadratic.
Efficient Approach: The idea is to use the slight modification of Sieve of Eratosthenes. All the prime numbers up to the maximum value can be computed and stored in an array along with its position. Clearly, when the prime numbers are stored in an array, the index at which the number is stored is the position of the number in the series. After this precomputation, the answer can be calculated in constant time.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
#define limit 10000000
using namespace std;
int position[limit + 1];
void sieve()
{
position[0] = -1, position[1] = -1;
int pos = 0;
for ( int i = 2; i <= limit; i++) {
if (position[i] == 0) {
position[i] = ++pos;
for ( int j = i * 2; j <= limit; j += i)
position[j] = -1;
}
}
}
int main()
{
sieve();
int n = 11;
cout << position[n];
return 0;
}
|
Java
class GFG{
static final int limit = 10000000 ;
static int []position = new int [limit + 1 ];
static void sieve()
{
position[ 0 ] = - 1 ;
position[ 1 ] = - 1 ;
int pos = 0 ;
for ( int i = 2 ; i <= limit; i++) {
if (position[i] == 0 ) {
position[i] = ++pos;
for ( int j = i * 2 ; j <= limit; j += i)
position[j] = - 1 ;
}
}
}
public static void main(String[] args)
{
sieve();
int n = 11 ;
System.out.print(position[n]);
}
}
|
Python3
limit = 1000000
position = [ 0 ] * (limit + 1 )
def sieve():
position[ 0 ] = - 1
position[ 1 ] = - 1
pos = 0
for i in range ( 2 , limit + 1 ):
if (position[i] = = 0 ):
pos + = 1
position[i] = pos
for j in range ( i * 2 , limit + 1 ,i):
position[j] = - 1
if __name__ = = "__main__" :
sieve()
n = 11
print (position[n])
|
C#
using System;
class GFG{
static readonly int limit = 1000000;
static int []position = new int [limit + 1];
static void sieve()
{
position[0] = -1;
position[1] = -1;
int pos = 0;
for ( int i = 2; i <= limit; i++) {
if (position[i] == 0) {
position[i] = ++pos;
for ( int j = i * 2; j <= limit; j += i)
position[j] = -1;
}
}
}
public static void Main(String[] args)
{
sieve();
int n = 11;
Console.Write(position[n]);
}
}
|
Javascript
<script>
var limit = 10000000
var position = Array(limit+1).fill(0);
function sieve()
{
position[0] = -1, position[1] = -1;
var pos = 0;
for ( var i = 2; i <= limit; i++)
{
if (position[i] == 0)
{
position[i] = ++pos;
for ( var j = i * 2; j <= limit; j += i)
position[j] = -1;
}
}
}
sieve();
var n = 11;
document.write( position[n]);
</script>
|
Time Complexity: O(limit2)
Auxiliary Space: O(limit)
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