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g1501_1600.s1575_count_all_possible_routes.readme.md Maven / Gradle / Ivy

1575\. Count All Possible Routes

Hard

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] <= 109
*   All integers in `locations` are **distinct**.
*   `0 <= start, finish < locations.length`
*   `1 <= fuel <= 200`