Difficulty : Easy
Pre-requisites : Two pointers
Problem : Calculate the minimum possible penalty for the game described.
The heart of the solution is the following idea:
Let’s say that all the cells with the value of Ai greater than some K are prohibited ones and we can’t use them. Then, we need to find a such way from 0 to N+1 that minimizes the maximal value of By where y is the index of some visited cell. Let’s call this minimal value FK
Let’s check all the possible values of K, namely from 0 to 32000. It is clear that while increasing the value of K some cells stop being prohibited, and therefore the value of FK will only decrease. This gives rise to the two-pointers solution:
Initially, all the cells are prohibited. We iterate our K from 0 to 32000. When K gets increased, some new cells with the value of A equal to K stop being prohibited. But then, we might be able to make some cells prohibited again, and this time, forever. More precisely, when there exists a path from 0 to N+1, we just pick some non-prohibited cell with currently the maximal value of B and if there will still be a path from 0 to N+1, then we delete it. Otherwise, we just stop the deletion, try to improve the answer and proceed the cycle. To check, whether there is a path, it is necessary to find the largest distance between any two adjacent currently non-prohibited cells and to compare it to K.
But how to make the cells prohibited/non-prohibited and to find all the distances? STL does well here. We can store the indexes of currently non-prohibited cells in the set, as well as pairwise differences between adjacent non-prohibited cells. In order to pick a non-prohibited cell with the maximal value of B, we can use the priority queue.
This gives us O(N log N) solution.