codility 青蛙跳石头

https://stackoverflow.com/questions/39881068/frog-jumps-across-a-river-with-stones

Note that the earliest time you can reach x = i can be expressed by the following recurrence relation:

shortest[i] = if A[i] = -1 then +inf 
              else max(A[i], min{shortest[j] | i - D <= j < i})

So first there is a simple O(ND) solution using only dynamic programming.

This can actually be reduced to O(N + D) using an efficient algorithm to maintain the mininum of shortest on the sliding window [i-D ... i] (using double-ended queue).