# Find the position of the given Prime Number

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 = 11Output:5Explanation:

The prime numbers are 2, 3, 5, 7, 11, 13, 17, ….

Therefore, the position of 11 in this series is 5.Input:N = 13Output:6

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**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++

`// C++ program to find the position` `// of the given prime number` `#include <bits/stdc++.h>` `#define limit 10000000` `using` `namespace` `std;` `int` `position[limit + 1];` `// Function to precompute the position` `// of every prime number using Sieve` `void` `sieve()` `{` ` ` `// 0 and 1 are not prime numbers` ` ` `position[0] = -1, position[1] = -1;` ` ` `// Variable to store the position` ` ` `int` `pos = 0;` ` ` `for` `(` `int` `i = 2; i <= limit; i++) {` ` ` `if` `(position[i] == 0) {` ` ` `// Incrementing the position for` ` ` `// every prime number` ` ` `position[i] = ++pos;` ` ` `for` `(` `int` `j = i * 2; j <= limit; j += i)` ` ` `position[j] = -1;` ` ` `}` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `sieve();` ` ` `int` `n = 11;` ` ` `cout << position[n];` ` ` `return` `0;` `}` |

## Java

`// Java program to find the position` `// of the given prime number` `class` `GFG{ ` ` ` `static` `final` `int` `limit = ` `10000000` `;` `static` `int` `[]position = ` `new` `int` `[limit + ` `1` `];` ` ` `// Function to precompute the position` `// of every prime number using Sieve` `static` `void` `sieve()` `{` ` ` `// 0 and 1 are not prime numbers` ` ` `position[` `0` `] = -` `1` `;` ` ` `position[` `1` `] = -` `1` `;` ` ` ` ` `// Variable to store the position` ` ` `int` `pos = ` `0` `;` ` ` `for` `(` `int` `i = ` `2` `; i <= limit; i++) {` ` ` `if` `(position[i] == ` `0` `) {` ` ` ` ` `// Incrementing the position for` ` ` `// every prime number` ` ` `position[i] = ++pos;` ` ` `for` `(` `int` `j = i * ` `2` `; j <= limit; j += i)` ` ` `position[j] = -` `1` `;` ` ` `}` ` ` `}` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `sieve();` ` ` ` ` `int` `n = ` `11` `;` ` ` `System.out.print(position[n]);` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program to find the position` `# of the given prime number` `limit ` `=` `1000000` `position ` `=` `[` `0` `]` `*` `(limit ` `+` `1` `)` ` ` `# Function to precompute the position` `# of every prime number using Sieve` `def` `sieve():` ` ` `# 0 and 1 are not prime numbers` ` ` `position[` `0` `] ` `=` `-` `1` ` ` `position[` `1` `] ` `=` `-` `1` ` ` ` ` `# Variable to store the position` ` ` `pos ` `=` `0` ` ` `for` `i ` `in` `range` `(` `2` `, limit ` `+` `1` `):` ` ` `if` `(position[i] ` `=` `=` `0` `):` ` ` ` ` `# Incrementing the position for` ` ` `# every prime number` ` ` `pos ` `+` `=` `1` ` ` `position[i] ` `=` `pos` ` ` `for` `j ` `in` `range` `( i ` `*` `2` `, limit ` `+` `1` `,i):` ` ` `position[j] ` `=` `-` `1` ` ` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `sieve()` ` ` ` ` `n ` `=` `11` ` ` `print` `(position[n])` ` ` `# This code is contributed by chitranayal` |

## C#

`// C# program to find the position` `// of the given prime number` `using` `System;` `class` `GFG{ ` ` ` `static` `readonly` `int` `limit = 1000000;` `static` `int` `[]position = ` `new` `int` `[limit + 1];` ` ` `// Function to precompute the position` `// of every prime number using Sieve` `static` `void` `sieve()` `{` ` ` `// 0 and 1 are not prime numbers` ` ` `position[0] = -1;` ` ` `position[1] = -1;` ` ` ` ` `// Variable to store the position` ` ` `int` `pos = 0;` ` ` `for` `(` `int` `i = 2; i <= limit; i++) {` ` ` `if` `(position[i] == 0) {` ` ` ` ` `// Incrementing the position for` ` ` `// every prime number` ` ` `position[i] = ++pos;` ` ` `for` `(` `int` `j = i * 2; j <= limit; j += i)` ` ` `position[j] = -1;` ` ` `}` ` ` `}` `}` ` ` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `sieve();` ` ` ` ` `int` `n = 11;` ` ` `Console.Write(position[n]);` `}` `}` ` ` `// This code is contributed by Princi Singh` |

## Javascript

`<script>` `// Javascript program to find the position` `// of the given prime number` `var` `limit = 10000000` `var` `position = Array(limit+1).fill(0);` `// Function to precompute the position` `// of every prime number using Sieve` `function` `sieve()` `{` ` ` `// 0 and 1 are not prime numbers` ` ` `position[0] = -1, position[1] = -1;` ` ` `// Variable to store the position` ` ` `var` `pos = 0;` ` ` `for` `(` `var` `i = 2; i <= limit; i++)` ` ` `{` ` ` `if` `(position[i] == 0)` ` ` `{` ` ` `// Incrementing the position for` ` ` `// every prime number` ` ` `position[i] = ++pos;` ` ` `for` `(` `var` `j = i * 2; j <= limit; j += i)` ` ` `position[j] = -1;` ` ` `}` ` ` `}` `}` `// Driver code` `sieve();` `var` `n = 11;` `document.write( position[n]);` `// This code is contributed by noob2000.` `</script>` |

**Output:**

5

Time Complexity: O(limit^{2})

Auxiliary Space: O(limit)