Time complexity of rotation array m times using temporary array

Question

I am new to asymptotic analysis, on solving the array rotation problem on geeksforgeeks the first solution provided was using a temporary array, I tried implementing this logic and found that the space complexity is O(m) where m is the number of times the array needs to be rotated. But the time complexity is given as O(n) where n is the size of the array. But I found that the time complexity should be O(n+m)

Input arr[] = [1, 2, 3, 4, 5, 6, 7], m = 2, n =7
1) Store d elements in a temp array 
 temp[] = [1, 2] // It takes m iterations to copy the array
2) Shift rest of the arr[]
 arr[] = [3, 4, 5, 6, 7, 6, 7] // n-m iteration for shifting
3) Store back the d elements
 arr[] = [3, 4, 5, 6, 7, 1, 2] // m iterations for storing the elements back

Here is how I calculated time complexity

=> m+(n-m)+m
=> O(n+m)

Can anyone help me to figure out what I am doing wrong?

Answer 1

$O(n+m)$ and $O(n)$ are the same. $m$ can never be more than $n$, so $n+m\leq 2n =O(n)$.