Decode the Coding Interview in JavaScript: Real-World Examples/

...

/

Feature #3: Plot and Select Path

Feature #3: Plot and Select Path

Implementing the "Plot and Select Path" feature for our "Uber" project.

We'll cover the following...

Description

After obtaining the closest drivers and calculating the cost of traveling on different roads, we need to build a functionality to select a path from the driver’s location to the user’s location. All the drivers have to pass through multiple checkpoints to reach the user’s location. Each road between checkpoints will have a cost, which we learned how to calculate in the previous lesson. It is possible that some of the k chosen drivers might not have a path to the user due to unavailability. Unavailability can occur due to a driver already being in a ride that has ended but not reached its location. In some cases, the driver can also get booked by another user and become unavailable. The driver that has the path to the user’s location with the minimum accumulated cost will be selected.

We’ll be given a city map GMap as an array of different checkpoints. Another array pathCosts, at each index, will represent the cost of traveling between the corresponding checkpoints in GMap. We are also given some drivers, where each drivers[i] represents a single driver node. We need to determine whether a path from the driver node drivers[i] to a user node exists or not. If the path exists, return the accumulated sum of the checkpoints between the two nodes. Otherwise, return -1.

In the above example,

  • GMap has the values [["a","b"],["b","c"],["a","e"],["d","e"]].

  • pathCosts has the values [12,23,26,18].

  • drivers has the values ["c", "d", "e", "f"].

  • user has a value "a".

After calculating the total cost of each driver’s route to the user, we’ll select that driver that has a path to the user with the lowest cost. Here, the driver f has no path to the user due to unavailability.

Solution

The main problem comes down to finding a path between two nodes, if it exists. If the path exists, return the cumulative sums along the path as the result. Given the problem, it seems that we need to track the nodes where we come from. DFS (Depth-First Search), also known as the backtracking algorithm, will be applicable in this case.

Here is how the implementation will take place:

  1. Build the graph using the city map array GMap.

  2. Assign the cost to each edge while building the graph.

  3. Once the graph is built, evaluate each driver’s path in the drivers array by searching for a path between the driver node and the user node. ...

class Solution {
  
  getTotalCost(GMap, pathCosts, drivers, user) {
    var city = {}
    // Step 1). build the city from the GMap
    for (var i = 0; i < GMap.length; i++) {
      var checkPoints = GMap[i]
      var sourceNode = checkPoints[0]
      var destNode = checkPoints[1]
      pathCost = pathCosts[i]
      if (!city.hasOwnProperty(sourceNode))
          city[sourceNode] = {}
      if (!city.hasOwnProperty(destNode))
          city[destNode] = {}
      city[sourceNode][destNode] = pathCost
      city[destNode][sourceNode] = pathCost
    }
    // Step 2). Evaluate each driver via bactracking (DFS)
    // by verifying if there exists a path from driver to user
    var results = new Array(drivers.length).fill(0)
    for (var i = 0; i < drivers.length; i++) {
      var driver = drivers[i]
      if (!city.hasOwnProperty(driver) || !city.hasOwnProperty(user))
          results[i] = -1.0
      else {
          var visited = new Set()
          results[i] = this.backtrackEvaluate(city, driver, user, 0, visited)
      }
    }
    return results;
  }
  backtrackEvaluate(city, currNode, targetNode, accSum, visited) {
      // mark the visit
      visited.add(currNode)
      var ret = -1.0
      var neighbors = city[currNode]
      if (neighbors.hasOwnProperty(targetNode))
          ret = accSum + neighbors[targetNode]
      else {
          for (pair in neighbors) {
              var nextNode = pair
              if (visited.hasOwnProperty(nextNode))
                  continue
              ret = this.backtrackEvaluate(city, nextNode, targetNode,
                      accSum + neighbors[pair], visited)
              if (ret != -1.0)
                  break
          }
      }
      // unmark the visit, for the next backtracking
      visited.delete(currNode)
      return ret
  }
}
// Driver code
var sol = new Solution()
var G_map = [["a","b"],["b","c"],["a","e"],["d","e"]]
var path_costs = [12.0,23.0,26.0,18.0]
var drivers = ["c", "d", "e", "f"]
var user = "a"
var all_path_costs = sol.getTotalCost(G_map, path_costs, drivers, user)
console.log("Total cost of all paths", all_path_costs)
Plot and Select Path

Complexity measures

Time
...
Access this course and 1400+ top-rated courses and projects.