Count All Possible Routes

This page explains Java solution to problem Count All Possible Routes using Dynamic Programming algorithm.

Problem Statement

You are given an array of distinct positive integers locations where locations[i] represents the position of city i. You are also given integers start, finish and fuel representing the starting city, ending city, and the initial amount of fuel you have, respectively.

At each step, if you are at city i, you can pick any city j such that j != i and 0

Notice that fuel cannot become negative at any point in time, and that you are allowed to visit any city more than once (including start and finish).

Return the count of all possible routes from start to finish.

Since the answer may be too large, return it modulo 10^9 + 7.

Input: locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5
Output: 4
Explanation: The following are all possible routes, each uses 5 units of fuel:
1 -> 3
1 -> 2 -> 3
1 -> 4 -> 3
1 -> 4 -> 2 -> 3

Input: locations = [4,3,1], start = 1, finish = 0, fuel = 6
Output: 5
Explanation: The following are all possible routes: 1 -> 0, used fuel = 1
1 -> 2 -> 0, used fuel = 5
1 -> 2 -> 1 -> 0, used fuel = 5
1 -> 0 -> 1 -> 0, used fuel = 3
1 -> 0 -> 1 -> 0 -> 1 -> 0, used fuel = 5

Input: locations = [5,2,1], start = 0, finish = 2, fuel = 3
Output: 0

Input: locations = [1,2,3], start = 0, finish = 2, fuel = 40
Output: 615088286
Explanation: The total number of possible routes is 2615088300. Taking this number modulo 10^⁹ + 7 gives us 615088286.

Solution

package com.vc.hard;

import java.util.Arrays;

class CountAllPossibleRoutes {
    private int MOD = (int)1e9 + 7;

    public int countRoutes(int[] locations, int start, int finish, int fuel) {
        int n = locations.length;
        /**
            Bottom up Dynamic Programming
            dp[i][j]: Number of way you can reach finish from ith position with j quantity of fuel
         */
        int[][] dp = new int[n][fuel + 1];

        /**
            We can reach to finish from finish with even zero quantity of fuel,
            so assign 1 for dp[finish][Any quantity of fuel]
         */
         Arrays.fill(dp[finish], 1);

        for(int startingFuel = 0; startingFuel <= fuel; startingFuel++) {
            for(int from = 0; from < n; from++) {
                for(int to = 0; to < n; to++) {
                    if(from == to) continue;
                    int requiredFuel = Math.abs(locations[from] - locations[to]);
                    if(startingFuel >= requiredFuel) {
                        dp[from][startingFuel] = (dp[from][startingFuel] + dp[to][startingFuel - requiredFuel]) % MOD;
                    }
                }
            }
        }
        return dp[start][fuel];
    }
}

Time Complexity

O(N² * Fuel) Where
N is total number of locations in an input array
Fuel is starting quantity of fuel

Space Complexity

O(N * Fuel) Where
N is total number of locations in an input array
Fuel is starting quantity of fuel