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Program For Closest Prime Number

Last Updated : 23 Jul, 2025
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Given a number N, you have to print its closest prime number. The prime number can be lesser, equal, or greater than the given number.

Condition: 1 ≤ N ≤ 100000

Examples:

Input : 16
Output: 17
Explanation: The two nearer prime number of 16 are 13 and 17. But among these, 
17 is the closest(As its distance is only 1(17-16) from the given number).

Input : 97
Output : 97
Explanation : The closest prime number in this case is the given number number 
itself as the distance is 0 (97-97).

Approach : 

  1. Using Sieve of Eratosthenes store all prime numbers in a Vector.
  2. Copy all elements in vector to the new array.
  3. Use the upper bound to find the upper bound of the given number in an array.
  4. As the array is already sorted in nature, compare previous and current indexed numbers in an array.
  5. Return number with the smallest difference.

Below is the implementation of the approach.

C++ 
#include <iostream>
#include <vector>
using namespace std;

const int MAX = 100005;

vector<int> primeNumber;

// Sieve of Eratosthenes algorithm to find all prime numbers up to MAX
void sieveOfEratosthenes() {
    // Create a boolean array "prime[0..n]" and initialize all entries as true.
    // A value in prime[i] will finally be false if i is not a prime, else true.
    bool prime[MAX + 1];
    for (int i = 0; i <= MAX; i++) {
        prime[i] = true;
    }

    // Update all multiples of p
    for (int p = 2; p * p <= MAX; p++) {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
            for (int i = p * p; i <= MAX; i += p) {
                prime[i] = false;
            }
        }
    }

    // Add all prime numbers to the vector
    for (int i = 2; i <= MAX; i++) {
        if (prime[i] == true) {
            primeNumber.push_back(i);
        }
    }
}

// Binary search to find the index of the smallest element greater than number
int upper_bound(vector<int> arr, int low, int high, int number) {
    // Base case
    if (low > high) {
        return low;
    }

    // Find the middle index
    int mid = low + (high - low) / 2;

    // If arr[mid] is less than or equal to number, search in the right subarray
    if (arr[mid] <= number) {
        return upper_bound(arr, mid + 1, high, number);
    }

    // If arr[mid] is greater than number, search in the left subarray
    return upper_bound(arr, low, mid - 1, number);
}

// Function to find the closest prime number to a given number
int closestPrime(int number) {
    // Handle special case of number 1 explicitly
    if (number == 1) {
        return 2;
    }
    else {
        // Generate all prime numbers using Sieve of Eratosthenes algorithm
        sieveOfEratosthenes();

        // Convert vector to array for binary search
        int n = primeNumber.size();
        int arr[n];
        for (int i = 0; i < n; i++) {
            arr[i] = primeNumber[i];
        }

        // Find the index of the smallest element greater than number
        int index = upper_bound(primeNumber, 0, n, number);

        // Check if the current element or the previous element is the closest
        if (arr[index] == number || arr[index - 1] == number) {
            return number;
        }
        else if (abs(arr[index] - number) < abs(arr[index - 1] - number)) {
            return arr[index];
        }
        else {
            return arr[index - 1];
        }
    }
}

// Driver program to test the above function
int main() {
    int number = 100;
    cout << closestPrime(number) << endl;
    return 0;
}
Java 
// Closest Prime Number in Java

import java.util.*;
import java.lang.*;

public class GFG {

    static int max = 100005;

    static Vector<Integer> primeNumber = new Vector<>();

    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.
        boolean prime[] = new boolean[max + 1];
        for (int i = 0; i <= max; 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
                for (int i = p * p; i <= max; i += p)
                    prime[i] = false;
            }
        }

        // Print all prime numbers
        for (int i = 2; i <= max; i++) {

            if (prime[i] == true)
                primeNumber.add(i);
        }
    }

    static int upper_bound(Integer arr[], int low, int high,
                           int X)
    {
        // Base Case
        if (low > high)
            return low;

        // Find the middle index
        int mid = low + (high - low) / 2;

        // If arr[mid] is less than
        // or equal to X search in
        // right subarray
        if (arr[mid] <= X) {
            return upper_bound(arr, mid + 1, high, X);
        }

        // If arr[mid] is greater than X
        // then search in left subarray
        return upper_bound(arr, low, mid - 1, X);
    }

    public static int closetPrime(int number)
    {

        // We will handle it (for number = 1) explicitly
        // as the lower/left number of 1 can give us
        // negative index which will cost Runtime Error.
        if (number == 1)
            return 2;
        else {

            // calling sieve of eratosthenes to
            // fill the array into prime numbers
            sieveOfEratosthenes();

            Integer[] arr = primeNumber.toArray(
                new Integer[primeNumber.size()]);
            // searching the index
            int index
                = upper_bound(arr, 0, arr.length, number);

            if (arr[index] == number
                || arr[index - 1] == number)
                return number;
            else if (Math.abs(arr[index] - number)
                     < Math.abs(arr[index - 1] - number))
                return arr[index];
            else
                return arr[index - 1];
        }
    }
    // Driver Program
    public static void main(String[] args)
    {
        int number = 100;
        System.out.println(closetPrime(number));
    }
}
Python3 
# python code for the above approach
import bisect
import math

MAX = 100005

prime_numbers = []

# Sieve of Eratosthenes algorithm to find all prime numbers up to MAX


def sieve_of_eratosthenes():
    # Create a boolean array "prime[0..n]" and initialize all entries as true.
    # A value in prime[i] will finally be false if i is not a prime, else true.
    prime = [True] * (MAX + 1)

    # Update all multiples of p
    for p in range(2, int(math.sqrt(MAX)) + 1):
        # If prime[p] is not changed, then it is a prime
        if prime[p]:
            for i in range(p * p, MAX + 1, p):
                prime[i] = False

    # Add all prime numbers to the list
    for i in range(2, MAX + 1):
        if prime[i]:
            prime_numbers.append(i)

# Function to find the closest prime number to a given number


def closest_prime(number):
    # Handle special case of number 1 explicitly
    if number == 1:
        return 2
    else:
        # Generate all prime numbers using Sieve of Eratosthenes algorithm
        sieve_of_eratosthenes()

        # Find the index of the smallest element greater than number
        index = bisect.bisect_left(prime_numbers, number)

        # Check if the current element or the previous element is the closest
        if prime_numbers[index] == number or prime_numbers[index - 1] == number:
            return number
        elif abs(prime_numbers[index] - number) < abs(prime_numbers[index - 1] - number):
            return prime_numbers[index]
        else:
            return prime_numbers[index - 1]


# Driver program to test the above function
if __name__ == '__main__':
    number = 100
    print(closest_prime(number))
JavaScript 
const MAX = 100005;
let primeNumber = [];

// Sieve of Eratosthenes algorithm to find all prime numbers up to MAX
function sieveOfEratosthenes() {
  // Create a boolean array "prime[0..n]" and initialize all entries as true.
  // A value in prime[i] will finally be false if i is not a prime, else true.
  let prime = Array(MAX + 1).fill(true);

  // Update all multiples of p
  for (let p = 2; p * p <= MAX; p++) {
    // If prime[p] is not changed, then it is a prime
    if (prime[p] == true) {
      for (let i = p * p; i <= MAX; i += p) {
        prime[i] = false;
      }
    }
  }

  // Add all prime numbers to the vector
  for (let i = 2; i <= MAX; i++) {
    if (prime[i] == true) {
      primeNumber.push(i);
    }
  }
}

// Binary search to find the index of the smallest element greater than number
function upper_bound(arr, low, high, number) {
  // Base case
  if (low > high) {
    return low;
  }

  // Find the middle index
  let mid = low + Math.floor((high - low) / 2);

  // If arr[mid] is less than or equal to number, search in the right subarray
  if (arr[mid] <= number) {
    return upper_bound(arr, mid + 1, high, number);
  }

  // If arr[mid] is greater than number, search in the left subarray
  return upper_bound(arr, low, mid - 1, number);
}

// Function to find the closest prime number to a given number
function closestPrime(number) {
  // Handle special case of number 1 explicitly
  if (number == 1) {
    return 2;
  } else {
    // Generate all prime numbers using Sieve of Eratosthenes algorithm
    sieveOfEratosthenes();

    // Find the index of the smallest element greater than number
    let index = upper_bound(primeNumber, 0, primeNumber.length, number);

    // Check if the current element or the previous element is the closest
    if (primeNumber[index] == number || primeNumber[index - 1] == number) {
      return number;
    } else if (Math.abs(primeNumber[index] - number) < Math.abs(primeNumber[index - 1] - number)) {
      return primeNumber[index];
    } else {
      return primeNumber[index - 1];
    }
  }
}

// Driver program to test the above function
let number = 100;
console.log(closestPrime(number));
C# 
//C# code for the above approach
using System;
using System.Collections.Generic;

public class Program
{
    const int MAX = 100005;
    static List<int> prime_numbers = new List<int>();

    // Sieve of Eratosthenes algorithm to find all prime numbers up to MAX
    static void sieve_of_eratosthenes()
    {
        // Create a boolean array "prime[0..n]" and initialize all entries as true.
        // 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 i = 0; i <= MAX; i++)
        {
            prime[i] = true;
        }

        // Update all multiples of p
        for (int p = 2; p <= Math.Sqrt(MAX); p++)
        {
            // If prime[p] is not changed, then it is a prime
            if (prime[p])
            {
                for (int i = p * p; i <= MAX; i += p)
                {
                    prime[i] = false;
                }
            }
        }

        // Add all prime numbers to the list
        for (int i = 2; i <= MAX; i++)
        {
            if (prime[i])
            {
                prime_numbers.Add(i);
            }
        }
    }

    // Function to find the closest prime number to a given number
    static int closest_prime(int number)
    {
        // Handle special case of number 1 explicitly
        if (number == 1)
        {
            return 2;
        }
        else
        {
            // Generate all prime numbers using Sieve of Eratosthenes algorithm
            sieve_of_eratosthenes();

            // Find the index of the smallest element greater than number
            int index = prime_numbers.BinarySearch(number);
            if (index < 0)
            {
                index = ~index;
            }

            // Check if the current element or the previous element is the closest
            if (prime_numbers[index] == number || prime_numbers[index - 1] == number)
            {
                return number;
            }
            else if (Math.Abs(prime_numbers[index] - number) < Math.Abs(prime_numbers[index - 1] - number))
            {
                return prime_numbers[index];
            }
            else
            {
                return prime_numbers[index - 1];
            }
        }
    }

    // Driver program to test the above function
    public static void Main()
    {
        int number = 100;
        Console.WriteLine(closest_prime(number));
    }
}

//This code is contributed by shivamsharma215

Output
101

Time Complexity: O(N log(log(N)))
Auxiliary Space: O(N)


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