Since we only need to know the final number of potatoes in each row after the updates, we can use a difference array of length ~N~ to keep track of the number of potatoes in each row. Each column can be updated in ~\mathcal{O}(1)~ time, making this part of the algorithm ~\mathcal{O}(M)~. After retrieving the actual values, we can use a two-pointer approach to find the minimum ~W~ required in ~\mathcal{O}(N)~ time. The total time complexity is therefore ~\mathcal{O}(N+M)~.