Minimize Y for given N to minimize difference between LCM and GCD

 Given an integer N, the task is to find a minimum possible Y that satisfies the following condition:

• Y > N.
• Y is not divisible by N (except N = 1).
• The absolute difference of gcd and lcm of N and Y i.e. | gcd(N, Y)?lcm(N, Y) | is as minimum as possible

Examples:

Input: N = 3 
Output: 4
?Explanation: When  Y = 4 is taken then GCD(3, 4) = 1 and LCM(3, 4) = 12, | 1?12 | = 11 . so, 11 is the smallest value we can get by choosing Y as 4.

Input: N = 4
?Output: 6

Approach: The problem can be solved based on the following idea:

• First we check N is prime. If it is true return N+1 .
• Else, find the smallestDivisor of N and print (smallestDivisor+1)*(N/smallestDivisor)

Follow the steps mentioned below to implement the above idea:

• If N is less than 1, return false.
• Iterate a loop from (i = 2) to square root of N and check if the number is prime or not. 
  • If yes then return false.
  • Otherwise, return true.
• If N is prime, then return N+1 as a minimized number Y.
  • Else find the smallest divisor of N.
  • Then return (smallestDivisor+1)*(N/smallestDivisor) as a minimized number Y.

Below is the implementation of the above approach.

C++

// C++ implementation
#include <bits/stdc++.h>
using namespace std;

// Function to find smallest divisor
int smallestdivisor(int n)
{
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return -1;
}
// Function to check if n 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;
}

void find(int n)
{
  if (isPrime(n)) {
    cout << (n + 1);
  }
  else {
    int k = smallestdivisor(n);
    cout << ((k + 1) * (n / k));
  }
}

int main()
{
  int N = 3;

  // Function Call
  find(N);
  //    cout << "GFG!";
  return 0;
}
// this code is contributed by ksam24000

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to find smallest divisor
    static long smallestdivisor(long n)
    {
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return i;
            }
        }
        return -1;
    }

    // Function to minimize number Y
    public static void find(long n)
    {
        if (isPrime(n)) {
            System.out.println(n + 1);
        }
        else {
            long k = smallestdivisor(n);
            System.out.println((k + 1) * (n / k));
        }
    }

    // Function to check if n is prime or not
    static boolean isPrime(long 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;
    }

    // Driver code
    public static void main(String[] args)
    {
        long N = 3;

        // Function Call
        find(N);
    }
}

Python3

# Python code to implement the approach

import math

# Function to find smallest divisor
def smallestdivisor(n):
    for i in range(2, int(math.sqrt(n)+1)):
        if(n % i == 0):
            return i

    return -1

# Function to minimize number Y
def find(n):
    if(isPrime(n)):
        print(n+1)
    else:
        k = smallestdivisor(n)
        print((k + 1) * (n/k))

# Function to check if n 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


N = 3
# Function call
find(N)

# This code is contributed by lokesh

C#

// C# code to implement the approach
using System;
public class GFG {

  // Function to find smallest divisor
  static int smallestdivisor(int n)
  {
    for (int i = 2; i <= Math.Sqrt(n); i++) {
      if (n % i == 0) {
        return i;
      }
    }
    return -1;
  }

  // Function to minimize number Y
  public static void find(int n)
  {
    if (isPrime(n)) {
      Console.WriteLine(n + 1);
    }
    else {
      int k = smallestdivisor(n);
      Console.WriteLine((k + 1) * (n / k));
    }
  }

  // Function to check if n is prime or not
  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;
  }

  // Driver code
  public static void Main(string[] args)
  {
    int N = 3;

    // Function Call
    find(N);
  }
}

// This code is contributed by AnkThon

Javascript

   // JavaScript code for the above approach

   // Function to find smallest divisor
   function smallestdivisor(n) {
     for (let i = 2; i <= Math.sqrt(n); i++) {
       if (n % i == 0) {
         return i;
       }
     }
     return -1;
   }

   // Function to minimize number Y
   function find(n) {
     if (isPrime(n)) {
      console.log(n + 1);
     }
     else {
       let k = smallestdivisor(n);
       console.log((k + 1) * (n / k));
     }
   }

   // Function to check if n is prime or not
   function isPrime(n) {
     // Corner case
     if (n < 1)
       return false;

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

     return true;
   }

   // Driver code

   let N = 3;

   // Function Call
   find(N);

// This code is contributed by Potta Lokesh

Output

4

Time Complexity: O(?N) // since there runs a loop from 0 to sqrt(n).
Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant