{ "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.
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
.
\n
Example 1:
\n\n\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\n2 <= locations.length <= 100
1 <= locations[i] <= 109
locations
are distinct.0 <= start, finish < locations.length
1 <= fuel <= 200
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 start
,finish
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
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。
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
)。
请你返回从 start
到 finish
所有可能路径的数目。
由于答案可能很大, 请将它对 10^9 + 7
取余后返回。
\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:
\n\n2 <= locations.length <= 100
1 <= locations[i] <= 109
locations
中的整数 互不相同 。0 <= start, finish < locations.length
1 <= fuel <= 200
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