Java Solution for Count All Possible Routes Problem
Intuition\n Describe your first thoughts on how to solve this problem. \n- We need to find the count of all possible routes from the start city to the finish ci
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- We need to find the count of all possible routes from the start city to the finish city, considering the fuel constraint.\n- At each step, we have the option to move from the current city to any other city, reducing the fuel by the distance between the cities.\n- We want to count the number of valid routes that reach the finish city with non-negative fuel.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. We use a 2D dynamic programming array dp to store the number of routes for each city and remaining fuel combination.\n1. The dimensions of the dp array are n (number of cities) and fuel + 1 (possible fuel amounts).\n1. We initialize the dp array with 0, except for the start city and the initial fuel, which is set to 1.\n1. We iterate over the remaining fuel from fuel to 0 and for each city, consider all possible next cities.\n1. If the current city and the next city are different, we calculate the fuel needed to move between them using Math.abs(locations[currCity] - locations[nextCity]).\n1. If the remaining fuel is greater than or equal to the fuel needed, we update the number of routes for the next city and remaining fuel by adding the count of routes for the current city and remaining fuel.\n1. Finally, we sum up the counts of all possible routes from the finish city using different remaining fuel amounts.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- The solution has three nested loops: one for the remaining fuel, one for the current city, and one for the next city.\n- Each loop iterates at most n times, where n is the number of cities.\n- Therefore, the time complexity of the solution is O(n^3).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n- The space complexity is O(n * fuel) since we are using a 2D array dp of size n x (fuel + 1) to store the number of routes.\n- As per the given constraints, the maximum number of cities can be 100 and the maximum fuel can be 200.\n- Hence, the space complexity is O(100 * 200), which can be considered constant or O(1) in practical terms.\n\n# Code\n```\nclass Solution \n{\n    public int countRoutes(int[] locations, int start, int finish, int fuel) \n    {\n        int MOD = 1000000007;\n        int n = locations.length;\n        int[][] dp = new int[n][fuel + 1];\n\n        // Base case: if start and finish are the same, it\'s a valid route\n        dp[start][fuel] = 1;\n\n        for (int remainingFuel = fuel; remainingFuel >= 0; remainingFuel--) \n        {\n            for (int currCity = 0; currCity < n; currCity++) \n            {\n                for (int nextCity = 0; nextCity < n; nextCity++) \n                {\n                    if (currCity != nextCity) \n                    {\n                        int fuelNeeded = Math.abs(locations[currCity] - locations[nextCity]);\n                        if (remainingFuel >= fuelNeeded) \n                        {\n                            dp[nextCity][remainingFuel - fuelNeeded] = (dp[nextCity][remainingFuel - fuelNeeded] + dp[currCity][remainingFuel]) % MOD;\n                        }\n                    }\n                }\n            }\n        }\n\n        int count = 0;\n        for (int i = 0; i <= fuel; i++) \n        {\n            count = (count + dp[finish][i]) % MOD;\n        }\n        return count;\n    }\n}\n```
['Java']