Minimum operations to reduce N to a prime number by subtracting with its highest divisor (original) (raw)

Last Updated : 11 Oct, 2025

Given a positive integer **N. In one operation subtract **N with its highest divisor other than **N and **1. The task is to find minimum operations required to reduce **N exactly to a prime number.

**Examples:

**Input: N = 38
**Output: 1
**Explanation: Highest divisor of 38 is 19, so subtract it (38 - 19) = 19. 19 is a prime number.
So, number of operations required = 1.

**Input: N = 69
**Output: 2

**Approach: This problem can be solved by using simple concepts of maths. Follow the steps below to solve the given problem.

At first, check whether **N is already prime or not.
- If **N is already a prime, return **0.
- Else, Initialize a variable say **count = 0 to store the number of operations required.
- Initialize a variable say i=2 and run a while loop till N!=i
  * Run while loop and on each iteration subtract current value of N with its largest divisor.
  * Compute the steps and increment the count by **1.
- Return the **count.

Below is the implementation of the above approach:

C++ `

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;

// Function check whether a number // is prime or not bool isPrime(int n) { // Corner case if (n <= 1) return false;

// Check from 2 to square root of n
for (int i = 2; i <= sqrt(n); i++)
    if (n % i == 0)
        return false;

return true;

}

// Function to minimum operations required // to reduce the number to a prime number int minOperation(int N) { // Because 1 cannot be converted to prime if (N == 1) return -1;

// If given number is already prime
// return 0
if (isPrime(N) == true) {
    return 0;
}

// If number is not prime
else {
    // Variable for total count
    int count = 0;
    int i = 2;

    // If number is not equal to i
    while (N != i) {
        // If N is completely divisible by i
        while (N % i == 0) {
            // Temporary variable to store
            // current number
            int temp = N;

            // Update the number by decrementing
            // with highest divisor
            N -= (temp / i);

            // Increment count by 1
            count++;
            if (isPrime(N))
                return count;
        }
        i++;
    }

    // Return the count
    return count;
}

}

// Driver Code int main() { int N = 38;

cout << minOperation(N);

return 0;

}

Java

// Java program for the above approach import java.util.Arrays;

class GFG { // Function check whether a number // is prime or not public static boolean isPrime(int n) { // Corner case if (n <= 1) return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= Math.sqrt(n); i++)
        if (n % i == 0)
            return false;

    return true;
}

// Function to minimum operations required
// to reduce the number to a prime number
public static int minOperation(int N) {
    // Because 1 cannot be converted to prime
    if (N == 1)
        return -1;

    // If given number is already prime
    // return 0
    if (isPrime(N) == true) {
        return 0;
    }

    // If number is not prime
    else {
        // Variable for total count
        int count = 0;
        int i = 2;

        // If number is not equal to i
        while (N != i) {
            // If N is completely divisible by i
            while (N % i == 0) {
                // Temporary variable to store
                // current number
                int temp = N;

                // Update the number by decrementing
                // with highest divisor
                N -= (temp / i);

                // Increment count by 1
                count++;
                if (isPrime(N))
                    return count;
            }
            i++;
        }

        // Return the count
        return count;
    }
}

// Driver Code
public static void main(String args[]) {
    int N = 38;

    System.out.println(minOperation(N));
}

}

// This code is contributed by gfgking.

Python

python program for the above approach

import math

Function check whether a number

is prime or not

def isPrime(n):

# Corner case
if (n <= 1):
    return False

# Check from 2 to square root of n
for i in range(2, int(math.sqrt(n)) + 1):
    if (n % i == 0):
        return False

return True

Function to minimum operations required

to reduce the number to a prime number

def minOperation(N):

# Because 1 cannot be converted to prime
if (N == 1):
    return -1

# If given number is already prime
# return 0
if (isPrime(N) == True):
    return 0

# If number is not prime
else:
    # Variable for total count
    count = 0
    i = 2

    # If number is not equal to i
    while (N != i):
      
        # If N is completely divisible by i
        while (N % i == 0):
          
            # Temporary variable to store
            # current number
            temp = N

            # Update the number by decrementing
            # with highest divisor
            N -= (temp // i)

            # Increment count by 1
            count += 1
            if (isPrime(N)):
                return count
        i += 1

    # Return the count
    return count

Driver Code

if name == "main": N = 38 print(minOperation(N))

This code is contributed by rakeshsahni

C#

// C# program for the above approach using System;

public class GFG {

// Function check whether a number
// is prime or not
public static bool isPrime(int n)
{
  
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= Math.Sqrt(n); i++)
        if (n % i == 0)
            return false;

    return true;
}

// Function to minimum operations required
// to reduce the number to a prime number
public static int minOperation(int N) {
    // Because 1 cannot be converted to prime
    if (N == 1)
        return -1;

    // If given number is already prime
    // return 0
    if (isPrime(N) == true) {
        return 0;
    }

    // If number is not prime
    else {
        // Variable for total count
        int count = 0;
        int i = 2;

        // If number is not equal to i
        while (N != i) {
            
            // If N is completely divisible by i
            while (N % i == 0) {
                
                // Temporary variable to store
                // current number
                int temp = N;

                // Update the number by decrementing
                // with highest divisor
                N -= (temp / i);

                // Increment count by 1
                count++;
                
                if (isPrime(N))
                    return count;
            }
            i++;
        }

        // Return the count
        return count;
    }
}

// Driver Code
public static void Main(string []args) {
    int N = 38;

    Console.WriteLine(minOperation(N));
}

}

// This code is contributed by AnkThon

JavaScript

**Time Complexity: O(sqrtN)
**Auxiliary Space: O(1)