# Time complexity of rotation array m times using temporary array

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?

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