Minimum value of maximum values in sub-segments … in O(n) complexity

Question

I interviewed with Amazon a few days ago. I could not answer one of the questions the asked me to their satisfaction. I have tried to get the answer after the interview but

Answer 1

@templatetypedef's answer works, but I think I have a more direct approach.

Start by computing the max for the following (closed) intervals:

[k-1, k-1]
[k-2, k-1]
[k-3, k-1]
...
[0, k-1]

Note that each of these can be computed in constant time from the preceeding one.

Next, compute the max for these intervals:

[k, k]
[k, k+1]
[k, k+2]
...
[k, 2k-1]

Now these intervals:

[2k-1, 2k-1]
[2k-2, 2k-1]
[2k-3, 2k-1]
...
[k+1, 2k-1]

Next you do the intervals from 2k to 3k-1 ("forwards intervals"), then from 3k-1 down to 2k+1 ("backwards intervals"). And so on until you reach the end of the array.

Put all of these into a big table. Note that each entry in this table took constant time to compute. Observe that there are at most 2*n intervals in the table (because each element appears once on the right side of a "forwards interval" and once on the left side of a "backwards interval").

Now, if [a,b] is any interval of width k, it must contain exactly one of 0, k, 2k, ...

Say it contains m*k.

Observe that the intervals [a, m*k-1] and [m*k ,b] are both somewhere in our table. So we can simply look up the max for each, and the max of those two values is the max of the interval [a,b].

So for any interval of width k, we can use our table to get its maximum in constant time. We can generate the table in O(n) time. Result follows.