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

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 opertation, 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++

 `#include ``using` `namespace` `std;`` ` `int` `minOperations(``int` `N, ``int` `M)``{``    ``    ``if` `(N == M && N == 0)``        ``return` `0;`` ` `    ``    ``if` `(N == M)``        ``return` `1;`` ` `    ``    ``if` `(N % 2 == M % 2)``        ``return` `2;`` ` `    ``    ``return` `-1;``}`` ` `int` `main()``{``    ``int` `N = 3;``    ``int` `M = 5;``    ``cout << minOperations(N, M);`` ` `    ``return` `0;``}`

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

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