{ "data": { "question": { "questionId": "1680", "questionFrontendId": "1575", "categoryTitle": "Algorithms", "boundTopicId": 400133, "title": "Count All Possible Routes", "titleSlug": "count-all-possible-routes", "content": "

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.

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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.

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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).

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Return the count of all possible routes from start to finish. Since the answer may be too large, return it modulo 109 + 7.

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\n

Example 1:

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\nInput: locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5\nOutput: 4\nExplanation: The following are all possible routes, each uses 5 units of fuel:\n1 -> 3\n1 -> 2 -> 3\n1 -> 4 -> 3\n1 -> 4 -> 2 -> 3\n
\n\n

Example 2:

\n\n
\nInput: locations = [4,3,1], start = 1, finish = 0, fuel = 6\nOutput: 5\nExplanation: The following are all possible routes:\n1 -> 0, used fuel = 1\n1 -> 2 -> 0, used fuel = 5\n1 -> 2 -> 1 -> 0, used fuel = 5\n1 -> 0 -> 1 -> 0, used fuel = 3\n1 -> 0 -> 1 -> 0 -> 1 -> 0, used fuel = 5\n
\n\n

Example 3:

\n\n
\nInput: locations = [5,2,1], start = 0, finish = 2, fuel = 3\nOutput: 0\nExplanation: It is impossible to get from 0 to 2 using only 3 units of fuel since the shortest route needs 4 units of fuel.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "统计所有可行路径", "translatedContent": "

给你一个 互不相同 的整数数组,其中 locations[i] 表示第 i 个城市的位置。同时给你 startfinish 和 fuel 分别表示出发城市、目的地城市和你初始拥有的汽油总量

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每一步中,如果你在城市 i ,你可以选择任意一个城市 j ,满足  j != i 且 0 <= j < locations.length ,并移动到城市 j 。从城市 i 移动到 j 消耗的汽油量为 |locations[i] - locations[j]||x| 表示 x 的绝对值。

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请注意, fuel 任何时刻都 不能 为负,且你 可以 经过任意城市超过一次(包括 start 和 finish )。

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请你返回从 start 到 finish 所有可能路径的数目。

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由于答案可能很大, 请将它对 10^9 + 7 取余后返回。

\n\n

 

\n\n

示例 1:

\n\n
\n输入:locations = [2,3,6,8,4], start = 1, finish = 3, fuel = 5\n输出:4\n解释:以下为所有可能路径,每一条都用了 5 单位的汽油:\n1 -> 3\n1 -> 2 -> 3\n1 -> 4 -> 3\n1 -> 4 -> 2 -> 3\n
\n\n

示例 2:

\n\n
\n输入:locations = [4,3,1], start = 1, finish = 0, fuel = 6\n输出:5\n解释:以下为所有可能的路径:\n1 -> 0,使用汽油量为 fuel = 1\n1 -> 2 -> 0,使用汽油量为 fuel = 5\n1 -> 2 -> 1 -> 0,使用汽油量为 fuel = 5\n1 -> 0 -> 1 -> 0,使用汽油量为 fuel = 3\n1 -> 0 -> 1 -> 0 -> 1 -> 0,使用汽油量为 fuel = 5\n
\n\n

示例 3:

\n\n
\n输入:locations = [5,2,1], start = 0, finish = 2, fuel = 3\n输出:0\n解释:没有办法只用 3 单位的汽油从 0 到达 2 。因为最短路径需要 4 单位的汽油。
\n\n

 

\n\n

提示:

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