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1575 - Count All Possible Routes (Hard)

https://leetcode.com/problems/count-all-possible-routes/

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 <= j < locations.length and move to city j. Moving from city i to city j reduces the amount of fuel you have by |locations[i] - locations[j]|. Please notice that |x| denotes the absolute value of x.

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 109 + 7.

Example 1:

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

Example 2:

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

Example 3:

Input: locations = [5,2,1], start = 0, finish = 2, fuel = 3
Output: 0
Explanation: It is impossible to get from 0 to 2 using only 3 units of fuel since the shortest route needs 4 units of fuel.

Constraints:

  • 2 <= locations.length <= 100
  • 1 <= locations[i] <= 10^9
  • All integers in locations are distinct.
  • 0 <= start, finish < locations.length
  • 1 <= fuel <= 200

Approach 1: Dynamic Programming​

Written by @wingkwong
class Solution {
public:
    int countRoutes(vector<int>& locations, int start, int finish, int fuel) {
        int M = 1e9 + 7, ans = 0, n = locations.size();
        // dp[i][j]: how many routes to reach finish from `i` using at most `j` fuel
        vector<vector<int>> dp(n, vector<int>(fuel + 1, -1));
        function<int(int,int)> dfs = [&](int from, int f) {
            // fuel < 0 - not possible -> return 0
            if (f < 0) return 0;
            // calculated before -> return it directly
            if (dp[from][f] != -1) return dp[from][f];
            // if the starting point is the finishing point, 
            // then there is one way
            int res = from == finish;
            // try all locations
            for (int to = 0; to < n; to++) {
                // skip the same point
                if (from == to) continue;
                // move to next position and 
                // substract the fuel
                (res += dfs(to, f - abs(locations[from] - locations[to]))) %= M;
            }
            // memoize the result
            return dp[from][f] = res;
        };
        return dfs(start, fuel);
    }
};