Minimum number of increment / decrements required to be performed on one of the two given numbers to make them non-coprime

Given two positive integers A and B, the task is to find the minimum number of increments/decrements required to be performed on either A or B to make both the numbers non-coprime.

Examples:

Input: A = 12, B = 3
Output: 0
Explanation:
As 12 & 3 are already non-coprimes, so the minimum count of increment/decrement operation required is 0.

Input: A = 7, B = 17
Output: 2

Approach: The given problem can be solved based on the following observations:

• If A and B have Greatest Common Divisor greater than 1 then no increment or decrement is to be performed, as numbers are already non-coprime.
• Now, check for the difference of 1 in both directions for A as well as B. Hence it requires only a single step to convert any number to an even number.
• If none of the above two cases applies then 2 increments/decrements operations are required to make the numbers A and B to their nearest even number so that the numbers become non-co primes.

Based on the above observations, follow the steps below to solve the problem:

• If the GCD of A and B is not equal to 1, then print 0 as no operation is required.
• Else if the GCD of one of the pair {{A + 1, B}, {A – 1, B}, {A, B + 1}, {A, B – 1}} is not equal to 1, then print 1 as only one operations is required.
• Otherwise, print 2.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to find the minimum number of
// increments/decrements operations required
// to make both the numbers non-coprime
int makeCoprime(int a, int b)
{
    // If a & b are already non-coprimes
    if (__gcd(a, b) != 1)
        return 0;

    // If a & b can become non-coprimes
    // by only incrementing/decrementing
    // a number only once
    if (__gcd(a - 1, b) != 1
        or __gcd(a + 1, b) != 1
        or __gcd(b - 1, a) != 1
        or __gcd(b + 1, a) != 1)
        return 1;

    // Otherwise
    return 2;
}

// Driver Code
int main()
{
    int A = 7, B = 17;
    cout << makeCoprime(A, B);

    return 0;
}

Java

// Java program for the above approach
public class GFG
{

  // function to calculate gcd
  static int __gcd(int a, int b)
  {

    // Everything divides 0
    if (a == 0)
      return b;
    if (b == 0)
      return a;

    // base case
    if (a == b)
      return a;

    // a is greater
    if (a > b)
      return __gcd(a-b, b);
    return __gcd(a, b-a);
  }

  // Function to find the minimum number of
  // increments/decrements operations required
  // to make both the numbers non-coprime
  static int makeCoprime(int a, int b)
  {

    // If a & b are already non-coprimes
    if (__gcd(a, b) != 1)
      return 0;

    // If a & b can become non-coprimes
    // by only incrementing/decrementing
    // a number only once
    if (__gcd(a - 1, b) != 1 || __gcd(a + 1, b) != 1
        || __gcd(b - 1, a) != 1
        || __gcd(b + 1, a) != 1)
      return 1;

    // Otherwise
    return 2;
  }

  // Driver code
  public static void main(String args[])
  {
    int A = 7, B = 17;
    System.out.println(makeCoprime(A, B));
  }
}

// This code is contributed by SoumikMondal

Python3

# Python3 program for the above approach
from math import gcd

# Function to find the minimum number of
# increments/decrements operations required
# to make both the numbers non-coprime
def makeCoprime(a, b):

    # If a & b are already non-coprimes
    if (gcd(a, b) != 1):
        return 0

    # If a & b can become non-coprimes
    # by only incrementing/decrementing
    # a number only once
    if (gcd(a - 1, b) != 1 or
        gcd(a + 1, b) != 1 or
        gcd(b - 1, a) != 1 or
        gcd(b + 1, a) != 1):
        return 1

    # Otherwise
    return 2

# Driver Code
if __name__ == '__main__':

    A = 7
    B = 17

    print(makeCoprime(A, B))

# This code is contributed by SURENDRA_GANGWAR

C#

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

class GFG{

// Function to calculate gcd
static int __gcd(int a, int b)
{

    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    // Base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return __gcd(a - b, b);

    return __gcd(a, b - a);
}

// Function to find the minimum number of
// increments/decrements operations required
// to make both the numbers non-coprime
static int makeCoprime(int a, int b)
{

    // If a & b are already non-coprimes
    if (__gcd(a, b) != 1)
        return 0;

    // If a & b can become non-coprimes
    // by only incrementing/decrementing
    // a number only once
    if (__gcd(a - 1, b) != 1 ||
        __gcd(a + 1, b) != 1 ||
        __gcd(b - 1, a) != 1 ||
        __gcd(b + 1, a) != 1)
        return 1;

    // Otherwise
    return 2;
}

// Driver Code
public static void Main(String[] args)
{
    int A = 7, B = 17;

    Console.Write(makeCoprime(A, B));
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

        // JavaScript program for the above approach

        function __gcd(a, b) {
            if (b == 0)
                return a;
            return __gcd(b, a % b);

        }

        // Function to find the minimum number of
        // increments/decrements operations required
        // to make both the numbers non-coprime
        function makeCoprime(a, b) {
            // If a & b are already non-coprimes
            if (__gcd(a, b) != 1)
                return 0;

            // If a & b can become non-coprimes
            // by only incrementing/decrementing
            // a number only once
            if (__gcd(a - 1, b) != 1
                || __gcd(a + 1, b) != 1
                || __gcd(b - 1, a) != 1
                || __gcd(b + 1, a) != 1)
                return 1;

            // Otherwise
            return 2;
        }

        // Driver Code

        let A = 7, B = 17;
        document.write(makeCoprime(A, B));

    // This code is contributed by Potta Lokesh

</script>

Output:

2

Time Complexity: O(log(A, B))
Auxiliary Space: O(1)