Minimum operations required to make two numbers equal

Given two integers A and B. the task is to find the minimum number of operations required to make A and B equal. In each operation, either of the below steps can be performed: 
 

• Increment either A or B with its initial value.
• Increment both A and B with their initial value

Examples: 
 

Input: A = 4, B = 10 
Output: 4 
Explanation: 
Initially A = 4, B = 10 
Operation 1: Increment A only: A = A + 4 = 8 
Operation 2: Increment A only: A = A + 4 = 12 
Operation 3: Increment A only: A = A + 4 = 16 
Operation 4: Increment A and B: A = A + 4 = 20 and B = B + 10 = 20 
They are equal now.
Input: A = 7, B = 23 
Output: 22 
Explanation: 
Initially A = 7, B = 23 
Operation 1 - 7: Increment A and B: A = 56 and B = 161 
Operation 8 - 22: Increment A: A = 161 and B = 161 
They are equal now. 
 

Approach: This problem can be solved using GCD. 
 

1. If A is greater than B, then swap A and B.
2. Now reduce B, such that gcd of A and B becomes 1.
3. Hence the minimum operations required to reach equal value is (B - 1).

For example: If A = 4, B = 10: 
 

• Step 1: Compare 4 and 10, as we always need B as the greater value. Here already B is greater than A. So, now no swap is required.
• Step 2: GCD(4, 10) = 2. So, we reduce B to B/2. Now A = 4 and B = 5. 
  GCD(4, 5) = 1, which was the target.
• Step 3: (Current value of B - 1) will be the required count. Here, Current B = 5. So (5 - 1 = 4), i.e. total 4 operations are required.

Below is the implementation of the above approach. 
 

// C++ program to find minimum
// operations required to
// make two numbers equal

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

// Function to return the
// minimum operations required
long long int minOperations(
    long long int A,
    long long int B)
{

    // Keeping B always greater
    if (A > B)
        swap(A, B);

    // Reduce B such that
    // gcd(A, B) becomes 1.
    B = B / __gcd(A, B);

    return B - 1;
}

// Driver code
int main()
{
    long long int A = 7, B = 15;

    cout << minOperations(A, B)
         << endl;

    return 0;
}

// Java program to find minimum
// operations required to
// make two numbers equal
class GFG{
 
// Function to return the
// minimum operations required
static int  minOperations(
    int  A,
    int  B)
{
 
    // Keeping B always greater
    if (A > B) {
        A = A+B;
        B = A-B;
        A = A-B;
    }
 
    // Reduce B such that
    // gcd(A, B) becomes 1.
    B = B / __gcd(A, B);
 
    return B - 1;
}
static int __gcd(int a, int b)  
{  
    return b == 0? a:__gcd(b, a % b);     
} 

// Driver code
public static void main(String[] args)
{
    int  A = 7, B = 15;
 
    System.out.print(minOperations(A, B)
         +"\n");
 
}
}

// This code contributed by sapnasingh4991

# Python program to find minimum 
# operations required to 
# make two numbers equal 
import math

# Function to return the 
# minimum operations required 
def minOperations(A, B):

    # Keeping B always greater 
    if (A > B):
        swap(A, B)

    # Reduce B such that 
    # gcd(A, B) becomes 1. 
    B = B // math.gcd(A, B); 

    return B - 1

# Driver code 
A = 7
B = 15

print(minOperations(A, B))

# This code is contributed by Sanjit_Prasad

// C# program to find minimum
// operations required to
// make two numbers equal
using System;

class GFG{
  
// Function to return the
// minimum operations required
static int  minOperations(
    int  A,
    int  B)
{
  
    // Keeping B always greater
    if (A > B) {
        A = A+B;
        B = A-B;
        A = A-B;
    }
  
    // Reduce B such that
    // gcd(A, B) becomes 1.
    B = B / __gcd(A, B);
  
    return B - 1;
}
static int __gcd(int a, int b)  
{  
    return b == 0? a:__gcd(b, a % b);     
} 
 
// Driver code
public static void Main(String[] args)
{
    int  A = 7, B = 15;
  
    Console.Write(minOperations(A, B)
         +"\n");
}
}

// This code is contributed by sapnasingh4991

<script>
// javascript program to find minimum
// operations required to
// make two numbers equal    
// Function to return the
    // minimum operations required
    function minOperations(A, B) 
{

        // Keeping B always greater
        if (A > B) {
            A = A + B;
            B = A - B;
            A = A - B;
        }

        // Reduce B such that
        // gcd(A, B) becomes 1.
        B = B / __gcd(A, B);

        return B - 1;
    }

    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }

    // Driver code
        var A = 7, B = 15;
        document.write(minOperations(A, B) + "\n");

// This code is contributed by Rajput-Ji 
</script>

14

Time Complexity: O(log(max(A, B))

Auxiliary Space: O(log(max(A, B))