Minimize operations to convert A to B by adding any odd integer or subtracting any even integer

Given two positive integers A and B. The task is to find the minimum number of operations required to convert number A into B. In a move, any one of the following operations can be applied on the number A:

• Select any odd integer x (x>0) and add it to A i.e. (A+x);
• Or, select any even integer y (y>0) and subtract it from A i.e. (A-y).

Examples:

Input: A = 2, B = 3
Output: 1
Explanation: Add odd x = 1 to A to obtain B (1 + 2 = 3).

Input: A = 7, B = 4
Output: 2
Explanation:  Two operations are required:
Subtract y = 4 from A (7 - 4 = 3).
Add x = 1 to get B (3 + 1 = 4).

Approach: This is an implementation-based problem. Follow the steps below to solve the given problem.

• Store absolute difference of A-B in variable diff.
• Check if A is equal to B. As the two integers are equal, the total number of operations will be 0.
• Else check if A < B.
  • If yes, check if their difference is odd or even.
    • If diff is odd, A+x operation is applied once (x is an odd integer equal to diff). The total number of operations is 1.
    • Else diff is even, apply A+x operation twice, Firstly where x is diff -1 (odd) and secondly where x is 1.  Or, A+x operation can be followed by A-y operation. In either case, the total number of operations will be 2.
• Else if A> B, the opposite set of operations are applied i.e.
  • If diff is even, A-y operation is applied once.
  • Else A-y operation can be applied followed by A+x operation. Or, A-y operation can be applied twice.

Hence, the number of operations will always be either 0, 1, or 2.

Below is the implementation for the above approach.

// C++ program for the given approach
#include <bits/stdc++.h>
using namespace std;

// Function to find
// minimum number of operations
int minOperations(int A, int B)
{
    // Variable to store
    // difference of A and B
    int diff;

    if (A == B)
        return 0;
    else if (A < B) {

        // A+x operation first
        diff = B - A;
        if (diff % 2 != 0)
            return 1;
        return 2;
    }
    else {

        // A-y operation first
        diff = A - B;
        if (diff % 2 == 0)
            return 1;
        return 2;
    }
}

// Driver code
int main()
{
    // Declaring integers A and B
    int A, B;

    // Initialising
    A = 7;
    B = 4;

    // Function call
    int ans = minOperations(A, B);

    // Displaying the result
    cout << ans;
    return 0;
}

// Java program for the given approach
import java.util.*;
class GFG{

// Function to find
// minimum number of operations
static int minOperations(int A, int B)
{
  
    // Variable to store
    // difference of A and B
    int diff;

    if (A == B)
        return 0;
    else if (A < B) {

        // A+x operation first
        diff = B - A;
        if (diff % 2 != 0)
            return 1;
        return 2;
    }
    else {

        // A-y operation first
        diff = A - B;
        if (diff % 2 == 0)
            return 1;
        return 2;
    }
}

// Driver code
public static void main(String[] args)
{
  
    // Declaring integers A and B
    int A, B;

    // Initialising
    A = 7;
    B = 4;

    // Function call
    int ans = minOperations(A, B);

    // Displaying the result
    System.out.print(ans);
}
}

// This code is contributed by 29AjayKumar 

# Python code for the above approach 

# Function to find
# minimum number of operations
def minOperations(A, B):

    # Variable to store
    # difference of A and B
    diff = None

    if (A == B):
        return 0;
    elif (A < B):

        # A+x operation first
        diff = B - A;
        if (diff % 2 != 0):
            return 1;
        return 2;
    else:

        # A-y operation first
        diff = A - B;
        if (diff % 2 == 0):
            return 1;
        return 2;
    
# Driver code

# Initialising A and B
A = 7;
B = 4;

# Function call
ans = minOperations(A, B);

# Displaying the result
print(ans);

# This code is contributed by gfgking

// C# program for the given approach
using System;
class GFG{

// Function to find
// minimum number of operations
static int minOperations(int A, int B)
{
  
    // Variable to store
    // difference of A and B
    int diff;

    if (A == B)
        return 0;
    else if (A < B) {

        // A+x operation first
        diff = B - A;
        if (diff % 2 != 0)
            return 1;
        return 2;
    }
    else {

        // A-y operation first
        diff = A - B;
        if (diff % 2 == 0)
            return 1;
        return 2;
    }
}

// Driver code
public static void Main()
{
  
    // Declaring integers A and B
    int A, B;

    // Initialising
    A = 7;
    B = 4;

    // Function call
    int ans = minOperations(A, B);

    // Displaying the result
    Console.Write(ans);
}
}

// This code is contributed by Samim Hossain Mondal.

  <script>
        // JavaScript code for the above approach 

        // Function to find
        // minimum number of operations
        function minOperations(A, B) 
        {
        
            // Variable to store
            // difference of A and B
            let diff;

            if (A == B)
                return 0;
            else if (A < B) {

                // A+x operation first
                diff = B - A;
                if (diff % 2 != 0)
                    return 1;
                return 2;
            }
            else {

                // A-y operation first
                diff = A - B;
                if (diff % 2 == 0)
                    return 1;
                return 2;
            }
        }

        // Driver code

        // Declaring integers A and B
        let A, B;

        // Initialising
        A = 7;
        B = 4;

        // Function call
        let ans = minOperations(A, B);

        // Displaying the result
        document.write(ans);

         // This code is contributed by Potta Lokesh
    </script>

2

Time Complexity: O(1)
Auxiliary Space: O(1)