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

  • Last Updated : 02 Aug, 2021

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)

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