Python ||  Python3 || Easy & Fast Solution
Approach\n Describe your first thoughts on how to solve this problem. \n\n- Define a recursive function, let\'s call it countRoutes, that takes the following pa
# Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n- Define a recursive function, let\'s call it countRoutes, that takes the following parameters: locations, start, finish, and fuel. This function will return the count of valid routes from the start location to the finish location with the given fuel.\n\n- Implement memoization to avoid redundant calculations and improve the performance of the recursive function. You can use the @lru_cache decorator to cache the function calls and their results.\n\n- Inside the countRoutes function, check for the base case: if the current fuel is less than zero, return 0 since there are no valid routes from that point.\n\n- Initialize a variable result to accumulate the count of routes. Start with 0.\n\n- Check if the current location is the finish location. If it is, increment result by 1 to count the route from the start to the finish.\n\n- Iterate over all possible destinations (j) from the current location (start), excluding the current location itself (start != j).\n\n- Calculate the remaining fuel after traveling from the current location to the next location (j). The remaining fuel can be calculated by subtracting the absolute difference between the x-coordinates and y-coordinates of the current and next locations: remaining_fuel = currFuel - (abs(locations[start][0] - locations[j][0]) + abs(locations[start][1] - locations[j][1])).\n\n- Make a recursive call to the countRoutes function with the next location (j) and the remaining fuel (remaining_fuel).\n\n- Update the result variable by adding the count of routes from the recursive call.\n\n- Return the result modulo 10^9 + 7 (result % (10 ** 9 + 7)) to keep the result within the given modulo value.\n\n- Outside the countRoutes function, call it with the initial values of start, finish, and fuel to get the final count of routes.\n\nThe recursive function will explore all possible routes from the start to the finish location, considering the remaining fuel at each step. The use of memoization will significantly improve the efficiency of the solution by avoiding redundant calculations.\n\n\n\n# Code\n```\nfrom typing import List\nfrom functools import lru_cache\n\nclass Solution:\n    def countRoutes(self, locations: List[int], start: int, finish: int, fuel: int) -> int:\n        mod = 10 ** 9 + 7\n        \n        # Define the recursive function with memoization\n        @lru_cache(None)\n        def dp(i, currFuel):\n            # Base case: If fuel is negative, no more valid routes can be taken\n            if currFuel < 0:\n                return 0\n            \n            result = 0\n            # If the current location is the finish, increment the result by 1\n            if i == finish:\n                result = 1\n            \n            # Iterate over all possible destinations from the current location\n            for j in range(len(locations)):\n                if i != j:\n                    # Calculate the remaining fuel after traveling to the next location\n                    remaining_fuel = currFuel - abs(locations[i] - locations[j])\n                    # Recursive call to calculate the number of routes from the next location\n                    result += dp(j, remaining_fuel)\n            \n            return result % mod\n\n        # Start the recursion from the start location with the given fuel\n        return dp(start, fuel)\n\n```
['Python3']