# Nearest prime less than given number n

Last Updated : 21 Mar, 2023

You are given a number n ( 3 <= n < 10^6 ) and you have to find nearest prime less than n?

Examples:

Input : n = 10
Output: 7

Input : n = 17
Output: 13

Input : n = 30
Output: 29

A simple solution for this problem is to iterate from n-1 to 2, and for every number, check if it is a prime. If prime, then return it and break the loop. This solution looks fine if there is only one query. But not efficient if there are multiple queries for different values of n.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include using namespace std;   // Function to return nearest prime number int prime(int n) {     // All prime numbers are odd except two     if (n & 1)         n -= 2;     else         n--;       int i, j;     for (i = n; i >= 2; i -= 2) {         if (i % 2 == 0)             continue;         for (j = 3; j <= sqrt(i); j += 2) {             if (i % j == 0)                 break;         }         if (j > sqrt(i))             return i;     }     // It will only be executed when n is 3     return 2; }   // Driver Code int main() {     int n = 17;     cout << prime(n);     return 0; }

## C

 // C program for the above approach #include #include   // Function to return nearest prime number int prime(int n) {       // All prime numbers are odd except two     if (n & 1)         n -= 2;     else         n--;       int i, j;     for (i = n; i >= 2; i -= 2) {         if (i % 2 == 0)             continue;         for (j = 3; j <= sqrt(i); j += 2) {             if (i % j == 0)                 break;         }         if (j > sqrt(i))             return i;     }       // It will only be executed when n is 3     return 2; }   // Driver Code int main() {     int n = 17;     printf("%d", prime(n));     return 0; }   // This code is contributed by Sania Kumari Gupta

## Java

 // Java program for the above approach   import java.io.*;   class GFG {       // Function to return nearest prime number     static int prime(int n)     {           // All prime numbers are odd except two         if (n % 2 != 0)             n -= 2;         else             n--;           int i, j;         for (i = n; i >= 2; i -= 2) {             if (i % 2 == 0)                 continue;             for (j = 3; j <= Math.sqrt(i); j += 2) {                 if (i % j == 0)                     break;             }             if (j > Math.sqrt(i))                 return i;         }           // It will only be executed when n is 3         return 2;     }       // Driver Code     public static void main(String[] args)     {         int n = 17;         System.out.print(prime(n));     } }   // This code is contributed by subham348.

## Python3

 # Python program for the above approach   # Function to return nearest prime number from math import floor, sqrt   def prime(n):           # All prime numbers are odd except two     if (n & 1):         n -= 2     else:         n -= 1           i,j = 0,3       for i in range(n, 2, -2):           if(i % 2 == 0):             continue           while(j <= floor(sqrt(i)) + 1):             if (i % j == 0):                 break             j += 2           if (j > floor(sqrt(i))):             return i       # It will only be executed when n is 3     return 2   # Driver Code n = 17   print(prime(n))   # This code is contributed by shinjanpatra

## C#

 // C# program for the above approach using System;   class GFG {       // Function to return nearest prime number     static int prime(int n)     {           // All prime numbers are odd except two         if (n % 2 != 0)             n -= 2;         else             n--;           int i, j;         for (i = n; i >= 2; i -= 2) {             if (i % 2 == 0)                 continue;             for (j = 3; j <= Math.Sqrt(i); j += 2) {                 if (i % j == 0)                     break;             }             if (j > Math.Sqrt(i))                 return i;         }           // It will only be executed when n is 3         return 2;     }       // Driver Code     public static void Main()     {         int n = 17;         Console.Write(prime(n));     } }   // This code is contributed by subham348.

## Javascript



Output

13

Time complexity:  O(n log n + log n) = O(n log n)
Space complexity: O(1)
An efficient solution for this problem is to generate all primes less than 10^6 using Sieve of Sundaram and store then in a array in increasing order. Now apply modified binary search to search nearest prime less than n.

## C++

 // C++ program to find the nearest prime to n. #include #define MAX 1000000 using namespace std;   // array to store all primes less than 10^6 vector primes;   // Utility function of Sieve of Sundaram void Sieve() {     int n = MAX;       // In general Sieve of Sundaram, produces primes     // smaller than (2*x + 2) for a number given     // number x     int nNew = sqrt(n);       // This array is used to separate numbers of the     // form i+j+2ij from others where  1 <= i <= j     int marked[n/2+500] = {0};       // eliminate indexes which does not produce primes     for (int i=1; i<=(nNew-1)/2; i++)         for (int j=(i*(i+1))<<1; j<=n/2; j=j+2*i+1)             marked[j] = 1;       // Since 2 is a prime number     primes.push_back(2);       // Remaining primes are of the form 2*i + 1 such     // that marked[i] is false.     for (int i=1; i<=n/2; i++)         if (marked[i] == 0)             primes.push_back(2*i + 1); }   // modified binary search to find nearest prime less than N int binarySearch(int left,int right,int n) {     if (left<=right)     {         int mid = (left + right)/2;           // base condition is, if we are reaching at left         // corner or right corner of primes[] array then         // return that corner element because before or         // after that we don't have any prime number in         // primes array         if (mid == 0 || mid == primes.size()-1)             return primes[mid];           // now if n is itself a prime so it will be present         // in primes array and here we have to find nearest         // prime less than n so we will return primes[mid-1]         if (primes[mid] == n)             return primes[mid-1];           // now if primes[mid]n that         // mean we reached at nearest prime         if (primes[mid] < n && primes[mid+1] > n)             return primes[mid];         if (n < primes[mid])             return binarySearch(left, mid-1, n);         else             return binarySearch(mid+1, right, n);     }     return 0; }   // Driver program to run the case int main() {     Sieve();     int n = 17;     cout << binarySearch(0, primes.size()-1, n);     return 0; }

## Java

 // Java program to find the nearest prime to n. import java.util.*;   class GFG {       static int MAX=1000000;   // array to store all primes less than 10^6 static ArrayList primes = new ArrayList();   // Utility function of Sieve of Sundaram static void Sieve() {     int n = MAX;       // In general Sieve of Sundaram, produces primes     // smaller than (2*x + 2) for a number given     // number x     int nNew = (int)Math.sqrt(n);       // This array is used to separate numbers of the     // form i+j+2ij from others where 1 <= i <= j     int[] marked = new int[n / 2 + 500];       // eliminate indexes which does not produce primes     for (int i = 1; i <= (nNew - 1) / 2; i++)         for (int j = (i * (i + 1)) << 1;                 j <= n / 2; j = j + 2 * i + 1)             marked[j] = 1;       // Since 2 is a prime number     primes.add(2);       // Remaining primes are of the form 2*i + 1 such     // that marked[i] is false.     for (int i = 1; i <= n / 2; i++)         if (marked[i] == 0)             primes.add(2 * i + 1); }   // modified binary search to find nearest prime less than N static int binarySearch(int left,int right,int n) {     if (left <= right)     {         int mid = (left + right) / 2;           // base condition is, if we are reaching at left         // corner or right corner of primes[] array then         // return that corner element because before or         // after that we don't have any prime number in         // primes array         if (mid == 0 || mid == primes.size() - 1)             return primes.get(mid);           // now if n is itself a prime so it will be present         // in primes array and here we have to find nearest         // prime less than n so we will return primes[mid-1]         if (primes.get(mid) == n)             return primes.get(mid - 1);           // now if primes[mid]n that         // mean we reached at nearest prime         if (primes.get(mid) < n && primes.get(mid + 1) > n)             return primes.get(mid);         if (n < primes.get(mid))             return binarySearch(left, mid - 1, n);         else             return binarySearch(mid + 1, right, n);     }     return 0; }   // Driver code public static void main (String[] args) {     Sieve();     int n = 17;     System.out.println(binarySearch(0,                         primes.size() - 1, n)); } }   // This code is contributed by mits

## Python3

 # Python3 program to find the nearest # prime to n. import math MAX = 10000;   # array to store all primes less # than 10^6 primes = [];   # Utility function of Sieve of Sundaram def Sieve():       n = MAX;       # In general Sieve of Sundaram, produces     # primes smaller than (2*x + 2) for a     # number given number x     nNew = int(math.sqrt(n));       # This array is used to separate numbers     # of the form i+j+2ij from others where     # 1 <= i <= j     marked = [0] * (int(n / 2 + 500));       # eliminate indexes which does not     # produce primes     for i in range(1, int((nNew - 1) / 2) + 1):         for j in range(((i * (i + 1)) << 1),                         (int(n / 2) + 1), (2 * i + 1)):             marked[j] = 1;       # Since 2 is a prime number     primes.append(2);       # Remaining primes are of the form     # 2*i + 1 such that marked[i] is false.     for i in range(1, int(n / 2) + 1):         if (marked[i] == 0):             primes.append(2 * i + 1);   # modified binary search to find nearest # prime less than N def binarySearch(left, right, n):     if (left <= right):         mid = int((left + right) / 2);           # base condition is, if we are reaching         # at left corner or right corner of         # primes[] array then return that corner         # element because before or after that         # we don't have any prime number in         # primes array         if (mid == 0 or mid == len(primes) - 1):             return primes[mid];           # now if n is itself a prime so it will         # be present in primes array and here         # we have to find nearest prime less than         # n so we will return primes[mid-1]         if (primes[mid] == n):             return primes[mid - 1];           # now if primes[mid]n         # that means we reached at nearest prime         if (primes[mid] < n and primes[mid + 1] > n):             return primes[mid];         if (n < primes[mid]):             return binarySearch(left, mid - 1, n);         else:             return binarySearch(mid + 1, right, n);       return 0;   # Driver Code Sieve(); n = 17; print(binarySearch(0, len(primes) - 1, n));       # This code is contributed by chandan_jnu

## C#

 // C# program to find the nearest prime to n. using System; using System.Collections; class GFG {       static int MAX = 1000000;   // array to store all primes less than 10^6 static ArrayList primes = new ArrayList();   // Utility function of Sieve of Sundaram static void Sieve() {     int n = MAX;       // In general Sieve of Sundaram, produces     // primes smaller than (2*x + 2) for a     // number given number x     int nNew = (int)Math.Sqrt(n);       // This array is used to separate numbers of the     // form i+j+2ij from others where 1 <= i <= j     int[] marked = new int[n / 2 + 500];       // eliminate indexes which does not produce primes     for (int i = 1; i <= (nNew - 1) / 2; i++)         for (int j = (i * (i + 1)) << 1;                  j <= n / 2; j = j + 2 * i + 1)             marked[j] = 1;       // Since 2 is a prime number     primes.Add(2);       // Remaining primes are of the form 2*i + 1     // such that marked[i] is false.     for (int i = 1; i <= n / 2; i++)         if (marked[i] == 0)             primes.Add(2 * i + 1); }   // modified binary search to find // nearest prime less than N static int binarySearch(int left, int right, int n) {     if (left <= right)     {         int mid = (left + right) / 2;           // base condition is, if we are reaching at left         // corner or right corner of primes[] array then         // return that corner element because before or         // after that we don't have any prime number in         // primes array         if (mid == 0 || mid == primes.Count - 1)             return (int)primes[mid];           // now if n is itself a prime so it will be         // present in primes array and here we have         // to find nearest prime less than n so we         // will return primes[mid-1]         if ((int)primes[mid] == n)             return (int)primes[mid - 1];           // now if primes[mid]n         // that mean we reached at nearest prime         if ((int)primes[mid] < n &&             (int)primes[mid + 1] > n)             return (int)primes[mid];         if (n < (int)primes[mid])             return binarySearch(left, mid - 1, n);         else             return binarySearch(mid + 1, right, n);     }     return 0; }   // Driver code static void Main() {     Sieve();     int n = 17;     Console.WriteLine(binarySearch(0,                       primes.Count - 1, n)); } }   // This code is contributed by chandan_jnu

## PHP

 n         // that means we reached at nearest prime         if (\$primes[\$mid] < \$n && \$primes[\$mid + 1] > \$n)             return \$primes[\$mid];         if (\$n < \$primes[\$mid])             return binarySearch(\$left, \$mid - 1, \$n);         else             return binarySearch(\$mid + 1, \$right, \$n);     }     return 0; }   // Driver Code Sieve(); \$n = 17; echo binarySearch(0, count(\$primes) - 1, \$n);       // This code is contributed by chandan_jnu ?>

## Javascript



Output

13

Time complexity: O(n log n)
Space complexity: O(n)

### Another Approach (Trial Division method) :

Start checking for prime numbers by dividing the given number n by all numbers less than n. The first prime number you encounter will be the nearest prime number less than n.

#### Algorithm:

1. Initialize a variable called “prime” to 0.
2. Starting from n-1, iterate through all numbers less than n in decreasing order.
3. For each number i, perform the following steps:
a. Initialize a variable called “is_prime” to true.
b. Starting from 2, iterate through all numbers less than i in increasing order.
c. For each number j, check if j divides i without leaving a remainder. If j does divide i, set “is_prime” to false and break out of the loop.
d. If “is_prime” is still true after checking all possible divisors, set “prime” to i and break out of the loop.
4. Return “prime” as the nearest prime less than n.

Below is the implementation of the above approach:

## C++

 // CPP code to find the nearest prime to N // Using Trial Division method #include #include using namespace std;   // Function to find the nearest prime to N // Using Trial Division method bool is_prime(int n) {     if (n <= 1) {         return false;     }     for (int i = 2; i <= sqrt(n); i++) {         if (n % i == 0) {             return false;         }     }     return true; }   int nearest_prime(int n) {     int prime = 0;     for (int i = n-1; i >= 2; i--) {         if (is_prime(i)) {             prime = i;             break;         }     }     return prime; }   int main() {     int n = 17;     int prime = nearest_prime(n);     if (prime == 0) {         cout << "There is no prime less than " << n << endl;     }     else {         cout << prime << endl;     }     return 0; }   // This code is contributed by Susobhan Akhuli

## Java

 // Java code to find the nearest prime to N // Using Trial Division method import java.util.*;   public class GFG {     // Function to find the nearest prime to N     // Using Trial Division method     public static boolean isPrime(int n)     {         if (n <= 1) {             return false;         }         for (int i = 2; i <= Math.sqrt(n); i++) {             if (n % i == 0) {                 return false;             }         }         return true;     }       public static int nearestPrime(int n)     {         int prime = 0;         for (int i = n - 1; i >= 2; i--) {             if (isPrime(i)) {                 prime = i;                 break;             }         }         return prime;     }       public static void main(String[] args)     {         int n = 17;         int prime = nearestPrime(n);         if (prime == 0) {             System.out.println(                 "There is no prime less than " + n);         }         else {             System.out.println(prime);         }     } }   // This code is contributed by Susobhan Akhuli

## Python3

 # Python code to find the nearest prime to N # using Trial Division method   import math   # Function to check if a number is prime or not def is_prime(n):     if n <= 1:         return False     for i in range(2, int(math.sqrt(n)) + 1):         if n % i == 0:             return False     return True   # Function to find the nearest prime to N # Using Trial Division method def nearest_prime(n):     prime = 0     for i in range(n - 1, 1, -1):         if is_prime(i):             prime = i             break     return prime     # Main function to test the above functions if __name__ == '__main__':     n = 17     prime = nearest_prime(n)     if prime == 0:         print(f"There is no prime less than {n}")     else:         print(prime)   # This code is contributed by sankar.

## C#

 // C# code implementation for the above approach   using System;   public class GFG {       // Function to check if a number is prime     static bool IsPrime(int n)     {         if (n <= 1) {             return false;         }         for (int i = 2; i <= Math.Sqrt(n); i++) {             if (n % i == 0) {                 return false;             }         }         return true;     }       // Function to find the nearest prime to n     static int NearestPrime(int n)     {         int prime = 0;         for (int i = n - 1; i >= 2; i--) {             if (IsPrime(i)) {                 prime = i;                 break;             }         }         return prime;     }       static public void Main()     {           // Code         int n = 17;         int prime = NearestPrime(n);         if (prime == 0) {             Console.WriteLine("There is no prime less than "                               + n);         }         else {             Console.WriteLine(prime);         }     } }   // This code is contributed by karthik.

## Javascript

 // Javascript code to find the nearest prime to N // Using Trial Division method   // Function to check if a number is prime function isPrime(n) {   if (n <= 1) {       return false;   }   for (let i = 2; i <= Math.sqrt(n); i++) {     if (n % i === 0) {       return false;     }   }   return true; }   // Function to find the nearest prime to n function nearestPrime(n) {   let prime = 0;   for (let i = n - 1; i >= 2; i--) {     if (isPrime(i)) {       prime = i;       break;     }   }   return prime; }   let n = 17; let prime = nearestPrime(n); if (prime === 0) {   console.log("There is no prime less than " + n); } else {   console.log(prime); }   // This code is contributed by karthik.

Output

13

Time Complexity: O(N* sqrt(N))
Auxiliary Space: O(1)

If you have another approach to solve this problem then please share in comments.

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