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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

14

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.