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1575. Count All Possible Routes

Description

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

Solutions

Solution 1: Memoization

We design a function $dfs(i,k)$, which represents the number of paths from city $i$ with $k$ remaining fuel to the destination $finish$. So the answer is $dfs(start,fuel)$.

The process of calculating the function $dfs(i,k)$ is as follows:

• If $k \lt

  locations[i] - locations[finish]

  $,thenreturn$0$.

• If $i=finish$, then the number of paths is $1$ at the beginning, otherwise it is $0$.

• Then, we traverse all cities $j$. If $j\ne i$, then we can move from city $i$ to city $j$, and the remaining fuel is $k -

  locations[i] - locations[j]

  $.Thenwecanaddthenumberofpathstotheanswer$dfs(j, k -

  locations[i] - locations[j]

  )$.

• Finally, we return the number of paths to the answer.

To avoid repeated calculations, we can use memoization.

The time complexity is $O(n^2\times m)$, and the space complexity is $O(n\times m)$. Where $n$ and $m$ are the size of the array $locations$ and $fuel$ respectively.

Solution 2: Dynamic Programming

We can also convert the memoization of solution 1 into dynamic programming.

We define $f[i][k]$ represents the number of paths from city $i$ with $k$ remaining fuel to the destination $finish$. So the answer is $f[start][fuel]$. Initially $f[finish][k]=1$, others are $0$.

Next, we enumerate the remaining fuel $k$ from small to large, and then enumerate all cities $i$. For each city $i$, we enumerate all cities $j$. If $j\ne i$, and $

locations[i] - locations[j]

\le k$,thenwecanmovefromcity$i$tocity$j$,andtheremainingfuelis$k -

locations[i] - locations[j]

$.Thenwecanaddthenumberofpathstotheanswer$f[j][k -

locations[i] - locations[j]

]$.

Finally, we return the number of paths to the answer $f[start][fuel]$.

The time complexity is $O(n^2\times m)$, and the space complexity is $O(n\times m)$. Where $n$ and $m$ are the size of the array $locations$ and $fuel$ respectively.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      private int[] locations;
      private int finish;
      private int n;
      private Integer[][] f;
      private final int mod = (int) 1e9 + 7;
  
      public int countRoutes(int[] locations, int start, int finish, int fuel) {
          n = locations.length;
          this.locations = locations;
          this.finish = finish;
          f = new Integer[n][fuel + 1];
          return dfs(start, fuel);
      }
  
      private int dfs(int i, int k) {
          if (k < Math.abs(locations[i] - locations[finish])) {
              return 0;
          }
          if (f[i][k] != null) {
              return f[i][k];
          }
          int ans = i == finish ? 1 : 0;
          for (int j = 0; j < n; ++j) {
              if (j != i) {
                  ans = (ans + dfs(j, k - Math.abs(locations[i] - locations[j]))) % mod;
              }
          }
          return f[i][k] = ans;
      }
  }
  

• class Solution {
  public:
      int countRoutes(vector<int>& locations, int start, int finish, int fuel) {
          int n = locations.size();
          int f[n][fuel + 1];
          memset(f, -1, sizeof(f));
          const int mod = 1e9 + 7;
          function<int(int, int)> dfs = [&](int i, int k) -> int {
              if (k < abs(locations[i] - locations[finish])) {
                  return 0;
              }
              if (f[i][k] != -1) {
                  return f[i][k];
              }
              int ans = i == finish;
              for (int j = 0; j < n; ++j) {
                  if (j != i) {
                      ans = (ans + dfs(j, k - abs(locations[i] - locations[j]))) % mod;
                  }
              }
              return f[i][k] = ans;
          };
          return dfs(start, fuel);
      }
  };
  

• class Solution:
      def countRoutes(
          self, locations: List[int], start: int, finish: int, fuel: int
      ) -> int:
          @cache
          def dfs(i: int, k: int) -> int:
              if k < abs(locations[i] - locations[finish]):
                  return 0
              ans = int(i == finish)
              for j, x in enumerate(locations):
                  if j != i:
                      ans = (ans + dfs(j, k - abs(locations[i] - x))) % mod
              return ans
  
          mod = 10**9 + 7
          return dfs(start, fuel)
  
  

• func countRoutes(locations []int, start int, finish int, fuel int) int {
  	n := len(locations)
  	f := make([][]int, n)
  	for i := range f {
  		f[i] = make([]int, fuel+1)
  		for j := range f[i] {
  			f[i][j] = -1
  		}
  	}
  	const mod = 1e9 + 7
  	var dfs func(int, int) int
  	dfs = func(i, k int) (ans int) {
  		if k < abs(locations[i]-locations[finish]) {
  			return 0
  		}
  		if f[i][k] != -1 {
  			return f[i][k]
  		}
  		if i == finish {
  			ans = 1
  		}
  		for j, x := range locations {
  			if j != i {
  				ans = (ans + dfs(j, k-abs(locations[i]-x))) % mod
  			}
  		}
  		f[i][k] = ans
  		return
  	}
  	return dfs(start, fuel)
  }
  
  func abs(x int) int {
  	if x < 0 {
  		return -x
  	}
  	return x
  }
  

• function countRoutes(locations: number[], start: number, finish: number, fuel: number): number {
      const n = locations.length;
      const f = Array.from({ length: n }, () => Array(fuel + 1).fill(-1));
      const mod = 1e9 + 7;
      const dfs = (i: number, k: number): number => {
          if (k < Math.abs(locations[i] - locations[finish])) {
              return 0;
          }
          if (f[i][k] !== -1) {
              return f[i][k];
          }
          let ans = i === finish ? 1 : 0;
          for (let j = 0; j < n; ++j) {
              if (j !== i) {
                  const x = Math.abs(locations[i] - locations[j]);
                  ans = (ans + dfs(j, k - x)) % mod;
              }
          }
          return (f[i][k] = ans);
      };
      return dfs(start, fuel);
  }
  
  

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