We are given $m$ arrays of $n$ elements each. For 90% of these arrays we run an $\mathcal O(n)$ algorithm and for the other 10% an $\mathcal O(n\log n)$ algorithm. What is the total time complexity? I think it is $\mathcal O(m n \log n)$ but I am not sure.
The O(n log n) eventually grows faster than the O(n) algorithm. We always take the fastest growing component of the algorithm when deciding the complexity, so you're right that the total complexity of the algorithm is O(m n log(n))
Ok, i dont speak English so i will be concise:
m * 90 / 100 = m'
m * 10 / 100 = m''
=> O(m'·n·log(n)) + O(m''·n) = O(m'·n·log(n) + m''·n)