Let us sort the drivers in descending points order and select the ~K^\text{th}~ driver in that sorted order. We need to determine whether (s)he can be the World Champion.

We can construct the best possible final race scenario for that driver. In such a race, (s)he would earn ~N~ points, while the current champion would earn only ~1~ point, the current runner-up only ~2~ points and so on. Generally, the driver in position ~p < K~ earns ~p~ points; drivers in positions ~p > K~ are irrelevant since they cannot have more points than driver ~K~ in the constructed scenario.

Can driver ~K~ be the World Champion in this scenario? The answer is 'yes' if (s)he has a sum greater than all other drivers, that is, if the relation

$$\displaystyle a[K]+N \ge \max\{a[1]+1, a[2]+2, \dots, a[K-1]+K-1\}$$

is satisfied, where ~a[p]~ is the number of points that driver ~p~ in the descending sequence had before the final race.

The solution that computes the maximum above (let us denote it by ~mx(K)~) for every ~K~ from scratch has complexity ~\mathcal O(N^2)~, which is too slow. Fortunately, the maximum is simple to compute gradually as needed: whenever the observed ~K~ is incremented, we can compare the previous maximum value with the current point value and update the maximum if needed. To formalize,

$$\displaystyle mx(K+1) = \max\{mx(K), a[K]+K\}$$

This reduces the complexity of finding the solution to ~\mathcal O(N)~. The total complexity is ~\mathcal O(N \log N)~ because of sorting.