Count All Possible Routes

class Solution {
public:
    int routeCounter(vector<int>& locs, int origin, int destination, int maxFuel) {
        int count = locs.size();
        vector<vector<int>> memo(count, vector<int>(maxFuel + 1));
        for(int f = 0; f <= maxFuel; f++) {
            memo[destination][f] = 1;
        }

        int result = 0;
        for (int fuel = 0; fuel <= maxFuel; fuel++) {
            for (int current = 0; current < count; current++) {
                for (int next = 0; next < count; next++) {
                    if (next == current) {
                        continue;
                    }
                    int cost = abs(locs[current] - locs[next]);
                    if (cost <= fuel) {
                        memo[current][fuel] = (memo[current][fuel] + memo[next][fuel - cost]) % 1000000007;
                    }
                }
            }
        }

        return memo[origin][maxFuel];
    }
};

The provided C++ solution introduces a method to count all possible routes from a given origin to a destination using a vehicle with a specified amount of fuel. The approach utilizes dynamic programming to solve the problem efficiently.

• Start by defining a routeCounter function that accepts a vector of locations, an origin index, a destination index, and the maximum fuel available.
• Establish a 2D memo vector where memo[i][j] represents the number of ways to reach location i with j units of fuel remaining.
• Initialize the memoization table by setting all ways to reach the destination with any fuel level to 1, since reaching the destination with any fuel left is a valid route.
• Use a triple nested loop to calculate the number of ways to reach each location with varying amounts of fuel:
  • The outer loop iterates over the fuel levels from 0 to maxFuel.
  • The second loop iterates over all possible current locations.
  • The inner loop iterates over all potential next locations, making sure to avoid the current location itself.
  • For each valid move (where the fuel cost to move to the next location does not exceed the current fuel level), update the memo table to add the count of possible ways to get to the next location after subtracting the fuel cost.
• Ensure results remain manageable by applying modulo 1000000007 to each computation.
• Ultimately, return memo[origin][maxFuel], which signifies the total number of ways to reach the destination from the origin with maxFuel units of fuel available based on the computed memo table.

This solution effectively uses memoization to break down the problem of counting routes into manageable subproblems, each representing a state defined by the current location and the remaining fuel. The computation of each state's value depends only on previously solved states, ensuring efficiency and correctness.

class Solution {
    public  int calculateRoutes(int[] locs, int strt, int end, int f) {
        int len = locs.length;
        int routes[][] = new int[len][f + 1];
        Arrays.fill(routes[end], 1);

        int result = 0;
        for (int j = 0; j <= f; j++) {
            for (int i = 0; i < len; i++) {
                for (int k = 0; k < len; k++) {
                    if (k == i) {
                        continue;
                    }
                    if (Math.abs(locs[i] - locs[k]) <= j) {
                        routes[i][j] = (routes[i][j] + routes[k][j - Math.abs(locs[i] - locs[k])]) %
                                       1000000007;
                    }
                }
            }
        }

        return routes[strt][f];
    }
}

The Java solution detailed above is designed to count all possible routes one can take from a starting point (strt) to a destination (end) given a number of locations and a fuel limit (f). This is implemented in the calculateRoutes method inside the Solution class. Here's a breakdown of the method:

• Start by determining the number of locations (len) and initialize a 2D array routes to store the interim results. The array dimensions are based on the number of locations and the fuel capacity plus one.

• Initially, set all possible fuel states at the destination (end) to 1 because reaching the destination with any remaining fuel is a valid scenario.

• Utilize a nested loop structure to compute the possibilities. The outermost loop iterates over each fuel state from 0 to f. For each fuel state:

  • Loop through each location i.
    • For each location i, consider each other location k.
      • Skip the iteration where k equals i.
      • Calculate the fuel required to move from location k to i. If it's within the available fuel (j), update routes[i][j]. The modulo operation ensures that the results stay within the limit of 1000000007 to avoid overflow issues.

• After populating the routes array, the solution for the given strt position and fuel f can be directly accessed and returned from routes[strt][f].

This approach efficiently handles the problem using dynamic programming by building off the number of routes from one location to another based on available fuel, which in turn saves significantly on computational resources and time compared to a naive recursive approach.

class PathFinder:
    def calculateRoutes(self, locs: List[int], init: int, goal: int, maxFuel: int) -> int:
        totalLocs = len(locs)
        route = [[0] * (maxFuel + 1) for _ in range(totalLocs)]

        for f in range(maxFuel + 1):
            route[goal][f] = 1

        for fuelLeft in range(maxFuel + 1):
            for currentPosition in range(totalLocs):
                for altPosition in range(totalLocs):
                    if altPosition == currentPosition:
                        continue
                    fuelCost = abs(locs[currentPosition] - locs[altPosition])
                    if fuelCost <= fuelLeft:
                        route[currentPosition][fuelLeft] = (route[currentPosition][fuelLeft] + route[altPosition][fuelLeft - fuelCost]) % 1000000007

        return route[init][maxFuel]

The solution provided calculates all possible routes from an initial location to a goal location within a given amount of fuel. The implementation uses dynamic programming in Python. Here is a breakdown of the solution:

• Define a class PathFinder with a method calculateRoutes.

• Variables:

  • locs: List of integers representing different locations.
  • init: Starting location index.
  • goal: Goal location index.
  • maxFuel: Maximum fuel available for the journey.

• An auxiliary matrix route is initialized to track the number of possible routes ending at each location with varying amounts of fuel left.

• The dynamic programming transition involves:

  1. Setting base cases where reaching the goal location with any fuel amount is always 1 possible route.
  2. Iterating over each amount of leftover fuel and each current position.
  3. For each current position, considering alternative positions that could be moved to by evaluating the fuel cost between them.
  4. Updating the count of routes to the current position based on reachable alternative positions and the fuel costs.

• Results are fetched from the route matrix for the initial location and the maximum fuel provided, giving the total possible routes.

The solution handles the problem efficiently using limited space and ensuring calculations are modular with a large constant to prevent overflow. This approach guarantees the accurate counting of routes under given constraints.