Problem Definition
This problem was asked by Square.
A competitive runner would like to create a route that starts and ends at his house, with the condition that the route goes entirely uphill at first, and then entirely downhill.
Given a dictionary of places of the form {location: elevation}, and a dictionary mapping paths between some of these locations to their corresponding distances, find the length of the shortest route satisfying the condition above. Assume the runner's home is location 0.
For example, suppose you are given the following input:
elevations = {0: 5, 1: 25, 2: 15, 3: 20, 4: 10}
paths = {
(0, 1): 10,
(0, 2): 8,
(0, 3): 15,
(1, 3): 12,
(2, 4): 10,
(3, 4): 5,
(3, 0): 17,
(4, 0): 10
}
In this case, the shortest valid path would be 0 -> 2 -> 4 -> 0, with a distance of 28.
Solution using BackTracking
To solve this problem we need to go through all possible valid paths and determine the minimum weight path.
It can be solved easily with Back Tracking using recursive method.
Solution Overview
- Start from the starting node which is 0.
- For each of the path that can be explored from current node, travel along that path if it is valid i.e. Elevation must be valid since we must travel strictly uphill and then downhill only. Anytime we encounter invalid node, we back track and follow different path. Also if node which has already been visited in given path is encountered we should back track since it will give cycle.
- If we can safely reach final destination node which is 0 again, we store the path summation in a list (OR update miniumum weight)
- After we have found a path, we need to back track again since there may be many possible paths and we need to find the optimal path.
Java Implementation
Following is implementation in Java. This prints lightest weight and the path. (GoLang version solution, below prints all valid paths with their weights.)
GoLang Implementation
Following is implementation in GoLang. This solution actually prints all valid paths with their path weight. It can be easily modified to return the minimum weight path.
Complexity
This solves the problem by traversing every possible paths. So the theoretical time complexity is $O(2^n)$.
However since the question has restriction that the path has to strictly ascend and then descend, we are returning path as invalid as soon as we encounter invalid path. Because of this restriction, in practice the time complexity will be $O(n).$
Alternative Solution with Memoization
In the previous solution, we start from first node 0 and keep on exploring all possible paths all the way to destination. During the exploration we keep updating track of path sum for all paths and update the best path.
In this solution, we compute the best path and best path weight that can be attained from given node. We cache this result and use it later when we encounter the same node in other path.
This solution does not require to pass
currentPath,
currentPathSum,
bestPath and
bestWeight to
findPathHelper()
Computation is done from the end. Idea is lightest tail path will be part of the solution path.
Following in implementation.
Complexity
The time complexity will be $O(n).$
