Number of continuous reductions of A from B or B from A to make them (1, 1)

Given two integers A and B, the task is to find the minimum number of operations required to change this pair to (1, 1). In each operation (A, B) can be changed to (A – B, B) where A > B.
Note: If there is no possible solution to reach (1, 1), print -1.

Examples:

Input: A = 7, B = 8
Output: 7
Explanation:
Operation 1: (A, B) => (7, 8) => (7, 1)
Operation 2: (A, B) => (7, 1) => (6, 1)
Operation 3: (A, B) => (6, 1) => (5, 1)
Operation 4: (A, B) => (5, 1) => (4, 1)
Operation 5: (A, B) => (4, 1) => (3, 1)
Operation 6: (A, B) => (3, 1) => (2, 1)
Operation 7: (A, B) => (2, 1) => (1, 1)

Input: A = 75, B = 17
Output: 10

Naive Approach: The idea is to use recursion and update the pair as (A, A – B), where A > B, and increase the number of operations required by 1. If at any step, any element of the pair is less than 1 then it is not possible to reach (1, 1).

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// minimum number of operations
// required to reach (1, 1)
#include <bits/stdc++.h>

using namespace std;

// Function to find the minimum
// number of steps required

int minimumSteps(int a, int b, int c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

        return -1;

    // Condition to check if the 

    // pair is reached to 1, 1

    if(a == 1 && b == 1)

        return c;

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a - b, b, c + 1);
} 
 
// Driver Code

int main()
{

    int a = 75;

    int b = 17;

    cout << minimumSteps(a, b, 0) << endl;
}
 
// This code is contributed by AbhiThakur

Java

// Java implementation to find the
// minimum number of operations
// required to reach (1, 1)

class GFG{

// Function to find the minimum
// number of steps required

static int minimumSteps(int a, int b, int c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

        return -1;

    // Condition to check if the 

    // pair is reached to 1, 1

    if(a == 1 && b == 1)

        return c;

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a - b, b, c + 1);
} 
 
// Driver Code

public static void main(String []args)
{

    int a = 75;

    int b = 17;

    System.out.println(minimumSteps(a, b, 0));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to find the
# minimum number of operations
# required to reach (1, 1)
 
# Function to find the minimum
# number of steps required

def minimumSteps(a, b, c):

    # Condition to check if it 

    # is not possible to reach

    if a < 1 or b < 1:

        return -1

    # Condition to check if the 

    # pair is reached to 1, 1

    if a == 1 and b == 1:

        return c

    # Condition to change the 

    # A as the maximum element

    if a < b:

        a, b = b, a

    return minimumSteps(a-b, b, c + 1)

# Driver Code

if __name__ == "__main__":

    a = 75; b = 17

    print(minimumSteps(a, b, 0))

// C# implementation to find the
// minimum number of operations
// required to reach (1, 1)

using System;
 
class GFG{

// Function to find the minimum
// number of steps required

static int minimumSteps(int a, int b, int c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

       return -1;

    // Condition to check if the 

    // pair is reached to 1, 1

    if(a == 1 && b == 1)

       return c;

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a - b, b, c + 1);
} 
 
// Driver Code

public static void Main()
{

    int a = 75;

    int b = 17;

    Console.WriteLine(minimumSteps(a, b, 0));
}
}
 
// This code is contributed by AbhiThakur

Javascript

<script>
 
// JavaScript implementation to find the
// minimum number of operations
// required to reach (1, 1)

// Function to find the minimum
// number of steps required

function minimumSteps(a, b, c)
{

    // Condition to check if it

    // is not possible to reach

    if(a < 1 || b < 1)

        return -1;

    // Condition to check if the

    // pair is reached to 1, 1

    if(a == 1 && b == 1)

        return c;

    // Condition to change the

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a - b, b, c + 1);
}
 
// Driver Code

    let a = 75;

    let b = 17;

    document.write(minimumSteps(a, b, 0) + "<br>");
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output:

Efficient Approach: The idea is to use the Euclidean algorithm to solve this problem. By this approach, we can go from (A, B) to (A % B, B) in the A/B steps. But if the minimum of the A and B is 1, then we can reach (1, 1) in A – 1 steps.

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// minimum number of operations
// required to reach (1, 1)
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the minimum

int minimumSteps(int a, int b, int c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

    {

        return -1;

    }

    // Condition to check if the 

    // pair is reached to 1, 1

    if(min(a, b) == 1)

    {

        return c + max(a, b) - 1;

    }

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        swap(a, b);

    }

    return minimumSteps(a % b, b, c + (a / b));
}
 
// Driver Code

int main()
{

    int a = 75, b = 17;

    cout << minimumSteps(a, b, 0) << endl;
 
    return 0;
}
 
// This code is contributed by rutvik_56

Java

// Java implementation to find the
// minimum number of operations
// required to reach (1, 1)

import java.util.*;

class GFG{
 
// Function to find the minimum

static int minimumSteps(int a, int b, int c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

    {

        return -1;

    }

    // Condition to check if the 

    // pair is reached to 1, 1

    if(Math.min(a, b) == 1)

    {

        return c + Math.max(a, b) - 1;

    }

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a % b, b, c + (a / b));
}
 
// Driver Code

public static void main(String[] args)
{

    int a = 75, b = 17;

    System.out.print(

           minimumSteps(a, b, 0) + "\n");
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to find the
# minimum number of operations
# required to reach (1, 1)
 
# Function to find the minimum
# number of steps required

def minimumSteps(a, b, c):

    # Condition to check if it 

    # is not possible to reach

    if a < 1 or b < 1:

        return -1

    # Condition to check if the 

    # pair is reached to 1, 1

    if min(a, b) == 1:

        return c + max(a, b) - 1

    # Condition to change the 

    # A as the maximum element

    if a < b:

        a, b = b, a

    return minimumSteps(a % b, b, c + a//b)

# Driver Code

if __name__ == "__main__":

    a = 75; b = 17

    print(minimumSteps(a, b, 0))

// C# implementation to find the
// minimum number of operations
// required to reach (1, 1)

using System;

class GFG{
 
// Function to find the minimum

static int minimumSteps(int a, int b, int c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

    {

        return -1;

    }

    // Condition to check if the 

    // pair is reached to 1, 1

    if(Math.Min(a, b) == 1)

    {

        return c + Math.Max(a, b) - 1;

    }

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a % b, b, c + (a / b));
}
 
// Driver Code

public static void Main()
{

    int a = 75, b = 17;

    Console.Write(minimumSteps(a, b, 0) + "\n");
}
}
 
// This code is contributed by Nidhi_biet

Javascript

<script>
 
// JavaScript implementation to find the
// minimum number of operations
// required to reach (1, 1)
 
// Function to find the minimum

function minimumSteps(a, b, c)
{

    // Condition to check if it 

    // is not possible to reach

    if(a < 1 || b < 1)

    {

        return -1;

    }

    // Condition to check if the 

    // pair is reached to 1, 1

    if(Math.min(a, b) == 1)

    {

        return c + Math.max(a, b) - 1;

    }

    // Condition to change the 

    // A as the maximum element

    if(a < b)

    {

        a = a + b;

        b = a - b;

        a = a - b;

    }

    return minimumSteps(a % b, b, c + parseInt(a / b));
}
 
// Driver Code

var a = 75, b = 17;

document.write(minimumSteps(a, b, 0) + "<br>");
 
</script>

Output:

Article Tags :

DSA

Mathematical

Recursion