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Minimize operations to convert (0, 0) to (N, M) by incrementing either or both by K

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  • Last Updated : 25 Jan, 2022

Given two integers N and M, the task is to calculate the minimum number of operations required to convert (0, 0) to (N, M) using the following operations:

  • Choose any integer K and convert (x, y) to (x + K, y + K).
  • Choose any integer K and convert (x, y) to (x – K, y + K) or (x + K, y – K).

Examples:

Input: N = 3, M = 5
Output: 2
Explanation: In 1st operation, take K = 4, and perform 1st operation i.e, (0 + 4, 0 + 4) -> (4, 4). In 2nd operation, take K = 1 and perform 2nd operation i.e, (4 – 1, 4 + 1) -> (3, 5) which is the required value. 

Input: N = 1, M = 4
Output: -1
Explanation: No possible sequence of given operations exists to convert (0, 0) to (1, 4). 

 

Approach: The given problem can be solved using the observation that each (N, M) pair can be divided into four following cases:

  • Case 1, where (N, M) = (0, 0). In such cases, 0 operations will be required.
  • Case 2, where N = M. In such cases, choose K = N and perform the 1st operation. Hence only one operation is required.
  • Case 3, where N and M are of the same parity, i.e, N % 2 = M % 2. In such cases, it can be observed that the required number of operations is always 2.
  • Case 4, where N and M are of different parity, i.e, N % 2 != M % 2. In such cases, no possible sequence of operations exists.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of operations required to convert
// a pair of integers (0, 0) to (N, M)
int minOperations(int N, int M)
{
    // Case 1
    if (N == M && N == 0)
        return 0;
 
    // Case 2
    if (N == M)
        return 1;
 
    // Case 3
    if (N % 2 == M % 2)
        return 2;
 
    // Not possible
    return -1;
}
 
// Driver Code
int main()
{
    int N = 3;
    int M = 5;
    cout << minOperations(N, M);
 
    return 0;
}


Java




// Java program to implement
// the above approach
class GFG {
 
  // Function to find the minimum number
  // of operations required to convert
  // a pair of integers (0, 0) to (N, M)
  static int minOperations(int N, int M) {
 
    // Case 1
    if (N == M && N == 0)
      return 0;
 
    // Case 2
    if (N == M)
      return 1;
 
    // Case 3
    if (N % 2 == M % 2)
      return 2;
 
    // Not possible
    return -1;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int N = 3;
    int M = 5;
    System.out.println(minOperations(N, M));
 
  }
}
 
// This code is contributed by Saurabh Jaiswal


Python3




# Python program of the above approach
 
# Function to find the minimum number
# of operations required to convert
# a pair of integers (0, 0) to (N, M)
def minOperations(N, M):
   
    # Case 1
    if N == M and N == 0:
        return 0
 
    # Case 2
    if N == M:
        return 1
 
    # Case 3
    if N % 2 == M % 2:
        return 2
 
    # Not possible
    return -1
 
# Driver Code
N = 3
M = 5
print(minOperations(N, M))
 
# This code is contributed by GFGking


C#




// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to find the minimum number
  // of operations required to convert
  // a pair of integers (0, 0) to (N, M)
  static int minOperations(int N, int M)
  {
     
    // Case 1
    if (N == M && N == 0)
      return 0;
 
    // Case 2
    if (N == M)
      return 1;
 
    // Case 3
    if (N % 2 == M % 2)
      return 2;
 
    // Not possible
    return -1;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 3;
    int M = 5;
    Console.Write(minOperations(N, M));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
    // JavaScript program of the above approach
 
    // Function to find the minimum number
    // of operations required to convert
    // a pair of integers (0, 0) to (N, M)
    const minOperations = (N, M) => {
        // Case 1
        if (N == M && N == 0)
            return 0;
 
        // Case 2
        if (N == M)
            return 1;
 
        // Case 3
        if (N % 2 == M % 2)
            return 2;
 
        // Not possible
        return -1;
    }
 
    // Driver Code
    let N = 3;
    let M = 5;
    document.write(minOperations(N, M));
 
// This code is contributed by rakeshsahni
 
</script>


 
 

Output

2

 

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

 


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