Minimum operations required to convert A into B using given conditions

Given two integers A and B, you have to choose two integers X and Y such that X must be odd, and Y must be even. Then you have to make A equal to B by using one out of two operations below at a time (possible zero time), the task is to find the minimum number of operations required to make integer A equal to B.

• Add X into A.
• Subtract Y from A.

Examples:

Input: A = 4, B = -5
Output: 2
Explanation: Two chosen integers are X = 1 and Y = 10, which are odd and even respectively.

• First operation: Add 1 into A, Then A = 4+1 = 5
• Second operation: Subtract 10 from A, then A = -10

Now A is equal to B. It can be verified that any other value of X and Y can't make A into B in less than 2 operations.

Input: A = 10, B = -12
Output: 1
Explanation: It can be verified that minimum operations will be 1, If we chose X and Y optimally.

Approach: Implement the idea below to solve the problem:

The problem is based on the observation. We can divide the following problem into sub-parts and solve it:

• If (A < B)
  • If difference between A and B is odd, then operations required will be 1.
  • If difference is multiple of 4, then operations required will be 3.
  • Otherwise, the minimum operations will be 2.
• If (A > B)
  • If difference between A and B is odd, then required operations will be 2.
  • Otherwise only 1 operation is required.
• If (A == B)
  • A is already equal to B, thus no operation is required. Answer will be 0.

Steps to solve the problem:

• If (A < B)
  • If ((B-A)%2 != 0), then minimum operations will be 1.
  • Else If ((B-A)%4 == 0), then minimum operations will be 3.
  • Else, then minimum operations will be 2.
• Else if (A > B)
  • If((A-B)%2 == 1), then minimum operations will be 2.
  • Else, minimum operation will be 1.
• Else If (A == B), then minimum operation is 0.

Below is the implementation of the above approah:

// C++ code to implement the approach

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

// Method to print the minimum
// number of operations
void Min_operations(int A, int B)
{

    if (A < B) {
        if ((B - A) % 2 != 0)
            cout << "1" << endl;
        else if ((B - A) % 4 == 0)
            cout << "3" << endl;
        else
            cout << "2" << endl;
    }
    else if (A > B) {
        if ((A - B) % 2 == 1)
            cout << "2" << endl;
        else
            cout << "1" << endl;
    }
    else {
        cout << "0" << endl;
    }
}

int main()
{

    // Inputs
    int A = 4;
    int B = -5;

    // Function call
    Min_operations(A, B);
}

// Java code to implement the approach

import java.util.*;

class Main {

    // Driver Function
    public static void main(String[] args)
        throws java.lang.Exception
    {

        // Inputs
        int A = 4;
        int B = -5;

        // Function call
        Min_operations(A, B);
    }

    // Method to print the minimum
    // number of operations
    public static void Min_operations(int A, int B)
    {

        if (A < B) {
            if ((B - A) % 2 != 0)
                System.out.println(1);
            else if ((B - A) % 4 == 0)
                System.out.println(3);
            else
                System.out.println(2);
        }
        else if (A > B) {
            if ((A - B) % 2 == 1)
                System.out.println(2);
            else
                System.out.println(1);
        }
        else {
            System.out.println(0);
        }
    }
}

# Python Implementation: 

# Method to print the minimum
# number of operations
def Min_operations(A, B):

    if A < B:
        if (B - A) % 2 != 0:
            print("1")
        elif (B - A) % 4 == 0:
            print("3")
        else:
            print("2")
    elif A > B:
        if (A - B) % 2 == 1:
            print("2")
        else:
            print("1")
    else:
        print("0")

# Inputs
A = 4
B = -5

# Function call
Min_operations(A, B)

# This code is contributed by Sakshi

using System;

public class GFG {

    // Method to print the minimum
    // number of operations
    public static void MinOperations(int A, int B)
    {
        if (A < B) {
            if ((B - A) % 2 != 0) {
                Console.WriteLine("1");
            }
            else if ((B - A) % 4 == 0) {
                Console.WriteLine("3");
            }
            else {
                Console.WriteLine("2");
            }
        }
        else if (A > B) {
            if ((A - B) % 2 == 1) {
                Console.WriteLine("2");
            }
            else {
                Console.WriteLine("1");
            }
        }
        else {
            Console.WriteLine("0");
        }
    }

    // Driver Function
    public static void Main()
    {
        // Inputs
        int A = 4;
        int B = -5;

        // Function call
        MinOperations(A, B);
    }
}
//This code is contributed by Rohit Singh

// JavaScript Implementation

// Method to print the minimum number of operations
function minOperations(A, B) {
  if (A < B) {
    if ((B - A) % 2 !== 0) {
      console.log("1");
    } else if ((B - A) % 4 === 0) {
      console.log("3");
    } else {
      console.log("2");
    }
  } else if (A > B) {
    if ((A - B) % 2 === 1) {
      console.log("2");
    } else {
      console.log("1");
    }
  } else {
    console.log("0");
  }
}

// Inputs
const A = 4;
const B = -5;

// Function call
minOperations(A, B);

// This code is contributed by Tapesh(tapeshdua420)

2

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