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
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
Solution
Great solution here
var countRoutes = function (locations, start, finish, fuel) {
// Map to keep track of visited nodes and paths to end with those conditions
let visited = new Map();
// Modulo to use
let mod = 10 ** 9 + 7;
// DFS function
let dfs = (current, fuel) => {
// Generate a key in the "2D" memo matrix
let id = `${current},${fuel}`;
// If we haven't visited already...
if (!visited.has(id)) {
let paths = 0;
// Count the paths; if we already finished, increment paths
if (current === finish) paths++;
// Go through all other positions...
for (let i = 0; i < locations.length; i++) {
if (i !== current) {
// Calculate remaining fuel
let remaining = fuel - Math.abs(locations[current] - locations[i]);
// If sufficient fuel, DFS further from current position
if (remaining >= 0) {
paths += dfs(i, remaining);
}
}
}
visited.set(id, paths % mod);
}
// Finally, cache result
return visited.get(id);
};
// Run DFS on starting node
return dfs(start, fuel);
};