How should I practice Optimal Transfer?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Superhuman online assessment problems in FastPrep.

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FastPrep tags this as a Superhuman online assessment coding practice problem when the problem catalog has that company and stage context.

Problem · Graph

Optimal Transfer

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● Hard

SuperhumanFULLTIMENEW GRADOA

Problem statement

There are n servers in a network, arranged in ascending order of their capacities. The capacity of server i is capacity[i], and all capacity values are distinct.

The distance between server i and server j is |capacity[i] - capacity[j]|. For each server, its closest server is the one with the smallest distance in capacity, and that closest server is guaranteed to be unique.

You may perform either of the following operations from a server x:

Connect directly to any server y at cost |capacity[x] - capacity[y]|.
Connect to the closest server of x at a fixed cost of 1.

Given query arrays fromServer and toServer, compute the minimum cost required to connect from fromServer[i] to toServer[i] for each query. A path may be direct or may route through intermediate servers.

Function

minimumTransferCost(capacity: int[], fromServer: int[], toServer: int[]) → int[]

Complete the function minimumTransferCost in the editor below.

minimumTransferCost has the following parameters:

int[] capacity: distinct server capacities in ascending order
int[] fromServer: starting server index for each query
int[] toServer: destination server index for each query

Returns

int[]: the minimum connection cost for each query in order

Examples

Example 1

capacity = [2, 7, 10]fromServer = [0, 1, 2]toServer = [2, 2, 1]return = [2, 1, 1]

The closest edges are 0 -> 1, 1 -> 2, and 2 -> 1, each costing 1. So the optimal costs are 0 -> 1 -> 2 = 2, 1 -> 2 = 1, and 2 -> 1 = 1.

Example 2

capacity = [1, 4, 9, 15]fromServer = [0, 3, 1]toServer = [3, 0, 3]return = [12, 3, 11]

The directed closest-server edges are 0 -> 1, 1 -> 0, 2 -> 1, and 3 -> 2. Therefore 0 -> 3 costs 1 + 5 + 6 = 12, 3 -> 0 costs 1 + 1 + 1 = 3, and 1 -> 3 costs 5 + 6 = 11.

Example 3

capacity = [5, 6, 20]fromServer = [2, 0]toServer = [0, 1]return = [2, 1]

Server 2 is closest to server 1, and server 1 is closest to server 0. So 2 -> 1 -> 0 costs 2, and 0 -> 1 costs 1.

Constraints

capacity is sorted in ascending order.
All values in capacity are distinct.
fromServer.length == toServer.length.