# Cheapest Bus Route - Google Top Interview Questions

### Problem Statement :

```You are given a two-dimensional list of integers connections.

Each element contains [f, t, id] meaning that bus id has a route from location f to location t.

It costs one unit of money to get on a new bus, but if you stay on the same bus continuously, you only pay one unit.

Return the minimum cost necessary to take the bus from location 0 to the largest location. Return -1 if it's not possible.

Constraints

n ≤ 1,000 where n is the number of bus stops

m ≤ 1,000 where m is the number of bus ids

c ≤ 100,000 where c is the length of connections

Example 1
Input

connections = [

[0, 1, 0],
[1, 2, 0],

[0, 3, 1],

[2, 4, 1],

[3, 0, 1]

]
Output

2

Explanation

We can get on bus 0 at location 0 and get out at location 2. Then we get on bus 1 to location 4.

Example 2

Input

connections = [

[0, 1, 0],

[1, 2, 1],

[2, 3, 0]

]

Output

3

Explanation

Even though we were on bus 0 originally, when we get on it again, we still pay a cost of one. This is
because the bus wasn't taken continuously.```

### Solution :

```                        ```Solution in C++ :

map<int, map<int, vector<int>>>
edges;           // edges[i][j] gives all locations k, for stop i and bus id j
int mindp[1005];     // mindp[i] = min cost from stop i
int dp[1005][1005];  // dp[i][j] = cost of being at station i with
bool minused[1005];  // have we tried transferring out of station i yet?
int solve(vector<vector<int>>& connections) {
int n = 0;
edges.clear();
int m = 0;
for (auto& e : connections) {
n = max(n, e[0] + 1);
n = max(n, e[1] + 1);
m = max(m, e[1] + 1);
edges[e[0]][e[2]].push_back(e[1]);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
dp[i][j] = 1e9;
}
}
for (int i = 0; i < n; i++) minused[i] = false;
mindp[0] = 0;
for (int i = 1; i < n; i++) mindp[i] = 1e9;
deque<pair<int, int>> q;
for (int i = 0; i < m; i++) {
dp[0][i] = 1;
q.emplace_back(0, i);
}
while (q.size()) {
int curr = q.front().first;
int currbus = q.front().second;
q.pop_front();
// stay on the same bus
for (int loc : edges[curr][currbus]) {
int cost = min(mindp[curr] + 1, dp[curr][currbus]);
if (cost < dp[loc][currbus]) {
dp[loc][currbus] = cost;
q.emplace_front(loc, currbus);
}
mindp[loc] = min(mindp[loc], dp[loc][currbus]);
}
if (!minused[curr]) {
// try all routes out
for (auto e : edges[curr]) {
int busid = e.first;
for (int loc : e.second) {
int cost = min(mindp[curr] + 1, dp[curr][busid]);
if (cost < dp[loc][busid]) {
dp[loc][busid] = cost;
q.emplace_back(loc, busid);
}
mindp[loc] = min(mindp[loc], dp[loc][busid]);
}
}
minused[curr] = true;
}
}
if (mindp[n - 1] == 1e9) return -1;
return mindp[n - 1];
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[][] connections) {
PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
return Integer.compare(a[1], b[1]);
}
});
// first element is the node, second element is shortest distance from 0 to that node
// last element(or third element) is BUS ID
Map<Integer, Set<State>> graph = new HashMap();
int maxy = -1;
int max_id = -1;
for (int[] con : connections) {
graph.putIfAbsent(con[0], new HashSet());

maxy = Math.max(maxy, Math.max(con[0], con[1]));
max_id = Math.max(max_id, con[2]);
}
// System.out.println("graph is " + graph.toString());
int[][] ref = new int[maxy + 1][max_id + 1];
for (int[] x : ref) Arrays.fill(x, Integer.MAX_VALUE);
for (int j = 0; j < ref[0].length; j++) {
ref[0][j] = 0;
}
boolean[][] vis = new boolean[maxy + 1][max_id + 1];

if (!graph.containsKey(0)) {
return -1;
}
for (State start : graph.get(0)) {
pq.offer(new int[] {0, 0, start.bus});
}
while (!pq.isEmpty()) {
int[] promising = pq.poll();
int node = promising[0];
int weight = promising[1];
int curr_bus = promising[2];
if (vis[node][curr_bus] || ref[node][curr_bus] < weight)
continue;
vis[node][curr_bus] = true;
if (!graph.containsKey(node))
continue;
for (State neighbor : graph.get(node)) {
int new_dist = weight;
if (neighbor.bus != curr_bus) {
new_dist++;
}
int new_node = neighbor.neighbor;
if (ref[new_node][neighbor.bus] > new_dist) {
pq.offer(new int[] {new_node, new_dist, neighbor.bus});
ref[new_node][neighbor.bus] = new_dist;
}
}
}
// System.out.println(Arrays.deepToString(ref));
int ret = Integer.MAX_VALUE;
for (int x : ref[maxy]) {
ret = Math.min(ret, x);
}
if (ret == Integer.MAX_VALUE)
return -1;
return ret + 1;
}
}
class State {
int neighbor;
int bus;
public State(int neighbor, int bus) {
this.neighbor = neighbor;
this.bus = bus;
}
@Override
public String toString() {
return neighbor + " " + bus;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, routes):
s = 0
target_end = float("-inf")

for route in routes:
target_end = max(target_end, route[0])
target_end = max(target_end, route[1])

if s == target_end:
return 0

graph = collections.defaultdict(list)

for start, end, id in routes:
graph[start].append((id, end))

pq = []

for id, end in graph[s]:
heapq.heappush(pq, (1, id, end))

visited = set()

while pq:
cost, id, e = heapq.heappop(pq)

if e == target_end:
return cost

for connected_id, end in graph[e]:
if (connected_id, end) in visited:
continue

if id == connected_id:
heapq.heappush(pq, (cost, connected_id, end))
else:
heapq.heappush(pq, (cost + 1, connected_id, end))

return -1```
```

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