Water collected between towers

Question

I recently came across an interview question asked by Amazon and I am not able to find an optimized algorithm to solve this question:

You are given an input array wh

Answer 1

None of the 17 answers already posted are really time-optimal.

For a single processor, a 2 sweep (left->right, followed by a right->left summation) is optimal, as many people have pointed out, but using many processors, it is possible to complete this task in O(log n) time. There are many ways to do this, so I'll explain one that is fairly close to the sequential algorithm.

Max-cached tree O(log n)

1: Create a binary tree of all towers such that each node contains the height of the highest tower in any of its children. Since the two leaves of any node can be computed independently, this can be done in O(log n) time with n cpu's.

2a: Then, for each node in the tree, starting at the root, let the right leaf have the value max(left, self, right). This will create the left-to-right monotonic sweep in O(log n) time, using n cpu's.

2b: To compute the right-to-left sweep, we do the same procedure as before. Starting with root of the max-cached tree, let the left leaf have the value max(left, self, right). These left-to-right (2a) and right-to-left (2b) sweeps can be done in parallel if you'd like to. They both use the max-cached tree as input, and generate one new tree each (or sets their own fields in original tree, if you prefer that).

3: Then, for each tower, the amount of water on it is min(ltr, rtl) - towerHeight, where ltr is the value for that tower in the left-to-right monotonic sweep we did before, i.e. the maximum height of any tower to the left of us (including ourselves¹), and rtl is the same for the right-to-left sweep.

4: Simply sum this up using a tree in O(log n) time using n cpu's, and we're done.

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¹ If the current tower is taller than all towers to the left of us, or taller than all towers to the the right of us, min(ltr, rtl) - towerHeight is zero.

Here's two other ways to do it.