Skip to content
Related Articles

Related Articles

Minimize moves to make X and Y equal using given subtraction operations

View Discussion
Improve Article
Save Article
  • Last Updated : 06 May, 2022
View Discussion
Improve Article
Save Article

Given two positive integers, X and Y, the task is to find the minimum number of operations to make X and y equal where in one operation, we can select any positive integer Z and do the following process.

  • If the number Z is even, subtract Z from X.
  • If the number Z is odd, add Z to X.

Examples:

Input: X = 4, Y = 7
Output: 1
Explanation:  Select Z = 3, then the new X will be 4 + 3 = 7. 
Hence, it requires only one operation.

Input: X = 6, Y = 6 
Output: 0
Explanation: Both of them are already the same. 
Hence, it requires 0 operations.

 

Approach: This problem can be solved using the Greedy approach based on the following observation:

There are only three possible answers: 

  • First, if X = Y, no operation is required,  
  • Second, If X > Y and X − Y is even or X < Y and Y − X is odd, then only 1 operation is required. 
  • Third, if X > Y and X − Y is odd or X < Y and Y − X is even, then there is need for 2 operations. One move extra is required for turning it to the second case

Follow the below steps to solve this problem:

  • Find the relation between X and Y.
  • Now based on the relation find how many moves are needed based on the above observation.

Below is the implementation of the above approach :

C++




// C++ code to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of moves required
int minMoves(int X, int Y)
{
    // Checking if both 'X' and 'Y'
    // are same or not.
    if (X == Y) {
        return 0;
    }
 
    // Checking if we can make X and Y equal
    // if we select Z = X - Y
    else if (X > Y && (X - Y) % 2 == 0) {
        return 1;
    }
 
    // Checking if we can make X and Y equal
    // if we select Z = Y - X
    else if (X < Y && (Y - X) % 2 == 1) {
        return 1;
    }
    else {
        return 2;
    }
}
 
// Driver Code
int main()
{
    int X = 4, Y = 7;
 
    // Function call
    cout << minMoves(X, Y);
    return 0;
}


Java




// Java code to implement the above approach
import java.util.*;
class GFG
{
   
  // Function to find the minimum number
  // of moves required
  public static int minMoves(int X, int Y)
  {
     
    // Checking if both 'X' and 'Y'
    // are same or not.
    if (X == Y) {
      return 0;
    }
 
    // Checking if we can make X and Y equal
    // if we select Z = X - Y
    else if (X > Y && (X - Y) % 2 == 0) {
      return 1;
    }
 
    // Checking if we can make X and Y equal
    // if we select Z = Y - X
    else if (X < Y && (Y - X) % 2 == 1) {
      return 1;
    }
    else {
      return 2;
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int X = 4, Y = 7;
 
    // Function call
    System.out.print(minMoves(X, Y));
  }
}
 
// This code is contributed by ashishsingh13122000.


Python3




# Python code to implement the above approach
 
# Function to find the minimum number
# of moves required
def minMoves(X, Y):
   
    # Checking if both 'X' and 'Y'
    # are same or not.
    if (X == Y):
        return 0
 
    # Checking if we can make X and Y equal
    # if we select Z = X - Y
    elif (X > Y and (X - Y) % 2 == 0):
        return 1
 
    # Checking if we can make X and Y equal
    # if we select Z = Y - X
    elif (X < Y and (Y - X) % 2 == 1):
        return 1
    else:
        return 2
 
# Driver Code
X, Y = 4, 7
 
# Function call
print(minMoves(X, Y))
 
# This code is contributed by shinjanpatra


C#




// C# code to implement the above approach
using System;
 
class GFG {
 
  // Function to find the minimum number
  // of moves required
  static int minMoves(int X, int Y)
  {
 
    // Checking if both 'X' and 'Y'
    // are same or not.
    if (X == Y) {
      return 0;
    }
 
    // Checking if we can make X and Y equal
    // if we select Z = X - Y
    else if (X > Y && (X - Y) % 2 == 0) {
      return 1;
    }
 
    // Checking if we can make X and Y equal
    // if we select Z = Y - X
    else if (X < Y && (Y - X) % 2 == 1) {
      return 1;
    }
    else {
      return 2;
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int X = 4, Y = 7;
 
    // Function call
    Console.Write(minMoves(X, Y));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
    // JavaScript code to implement the above approach
 
    // Function to find the minimum number
    // of moves required
    const minMoves = (X, Y) => {
     
        // Checking if both 'X' and 'Y'
        // are same or not.
        if (X == Y) {
            return 0;
        }
 
        // Checking if we can make X and Y equal
        // if we select Z = X - Y
        else if (X > Y && (X - Y) % 2 == 0) {
            return 1;
        }
 
        // Checking if we can make X and Y equal
        // if we select Z = Y - X
        else if (X < Y && (Y - X) % 2 == 1) {
            return 1;
        }
        else {
            return 2;
        }
    }
 
    // Driver Code
    let X = 4, Y = 7;
 
    // Function call
    document.write(minMoves(X, Y));
 
// This code is contributed by rakeshsahni
 
</script>


Output

1

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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!