# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

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

 `// 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 `

## Python3

 `# 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#

 `// 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 `

Output:

```14
```

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

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