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 10^9 + 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's impossible to get from 0 to 2 using only 3 units of fuel since the shortest route needs 4 units of fuel.
Example 4:
Input: locations = [2,1,5], start = 0, finish = 0, fuel = 3 Output: 2 Explanation: There are two possible routes, 0 and 0 -> 1 -> 0.
Example 5:
Input: locations = [1,2,3], start = 0, finish = 2, fuel = 40 Output: 615088286 Explanation: The total number of possible routes is 2615088300. Taking this number modulo 10^9 + 7 gives us 615088286.
Constraints:
2 <= locations.length <= 1001 <= locations[i] <= 10^9- All integers in
locationsare distinct. 0 <= start, finish < locations.length1 <= fuel <= 200
Related Topics:
Dynamic Programming
Let dp[i][j] be the count of all possible routes from city i with j fuel to city finish.
The initial state is dp[start][fuel].
Assume we can go from city i to city k, dp[i][j] += dp[k][j - dist(i, k)].
When i == finish and j >= 0, we should add 1 to dp[i][j].
dp[i][j] = sum( dp[k][j - dist[i, k]] | j - dist[i, k] >= 0 )
+ (i == start ? 1 : 0)
dp[start][fuel] is the answer.
// OJ: https://leetcode.com/problems/count-all-possible-routes/
// Author: github.com/lzl124631x
// Time: O(N^2 * F)
// Space: O(NF)
class Solution {
long dp[101][201] = {}, mod = 1e9+7, finish;
int dist(vector<int> &A, int a, int b) {
return abs(A[a] - A[b]);
}
int dfs(vector<int> &A, int start, int fuel) {
if (dp[start][fuel]) return dp[start][fuel];
long ans = 0;
if (start == finish) ans++;
for (int i = 0; i < A.size(); ++i) {
if (i == start || dist(A, start, i) + dist(A, i, finish) > fuel) continue;
ans = (ans + dfs(A, i, fuel - dist(A, i, start))) % mod;
}
return dp[start][fuel] = ans;
}
public:
int countRoutes(vector<int>& A, int start, int finish, int fuel) {
this->finish = finish;
return dfs(A, start, fuel);
}
};left[i][f]: the count of routes whose last move is leftwards to location i with f fuel
right[i][f]: the count of routes whose last move is rightwards to location i with f fuel
left[j][f] = left[j+1][f-d(j, j+1)] + right[j+1][f-d(j, j+1)] // go to j + 1 first then j
+ left[j+2][f-d(j, j+2)] + right[j+2][f-d(j, j+2)] // go to j + 2 first then j
+ ...
+ left[j+k][f-d(j, j+k)] + right[j+k][f-d(j, j+2)] // go to j + k first then j
We can apply this equation recursively so that all the left are replaced with right in the right side of the equation.
left[j][f] = right[j+1][f-d(j, j+1)]
+ 2 * right[j+2][f-d(j, j+2)]
+ ...
+ 2^(k-1) * right[j+k][f-d(j, j+k)]
In the righthand-side, the 2nd to last terms can be expressed as 2 * left[j+1][f-d(j, j+1)] because
left[j+1][f] = right[j+2][f-d(j+1, j+2)]
+ 2 * right[j+3][f-d(j+1, j+3)]
+ ...
+ 2^(k-2) * right[j+1+k-1][f-d(j+1, j+1+k-1)]
left[j+1][f-d(j, j+1)] = right[j+2][f-d(j, j+2)]
+ 2 * right[j+3][f-d(j, j+3)]
+ ...
+ 2^(k-2) * right[j+k][f-d(j, j+k)]
So we have this equation
left[j][f] = right[j+1][f-d(j, j+1)] + 2 * left[j+1][f-d(j, j+1)]
Symmetrically, we have
right[j][f] = left[j-1][f-d(j, j-1)] + 2 * right[j-1][f-d(j, j-1)]
// OJ: https://leetcode.com/problems/count-all-possible-routes/
// Author: github.com/lzl124631x
// Time: O(NlogN + NF)
// Space: O(NF)
class Solution {
public:
int countRoutes(vector<int>& A, int start, int finish, int fuel) {
int mod = 1e9+7, s = A[start], f = A[finish];
sort(begin(A), end(A));
start = lower_bound(begin(A), end(A), s) - begin(A);
finish = lower_bound(begin(A), end(A), f) - begin(A);
vector<vector<int>> left(A.size(), vector<int>(fuel + 1));
vector<vector<int>> right(A.size(), vector<int>(fuel + 1));
for (int f = 1; f <= fuel; ++f) {
for (int j = 0; j < A.size() - 1; ++j) {
int d = A[j + 1] - A[j];
if (f > d) left[j][f] = (right[j + 1][f - d] + 2 * left[j + 1][f - d] % mod) % mod;
else if (f == d) left[j][f] = j + 1 == start;
}
for (int j = 1; j < A.size(); ++j) {
int d = A[j] - A[j - 1];
if (f > d) right[j][f] = (left[j - 1][f - d] + 2 * right[j - 1][f - d] % mod) % mod;
else if (f == d) right[j][f] = j - 1 == start;
}
}
int ans = start == finish;
for (int f = 1; f <= fuel; ++f) ans = ((ans + left[finish][f]) % mod + right[finish][f]) % mod;
return ans;
}
};