# Cheapest Flights Within K Stops

## 1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?
• When does a given path given priority over another?

• In a basic Dijkstra algorithm, the only basis upon which one path is given priority over another is the cost. In this problem however, there is an additional constraint; length of route or (number of stops).
• What do we need to keep track of during the traversal?

• We need to keep track of all routes to the node and compare on the basis of (cost and number of stops).
``````  HAPPY CASE
Input: n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
Output: 700

Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
Output: 200

EDGE CASE
Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
Output: 500``````

## 2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

``````For graph problems, some things we want to consider are:
``````
• Dijkstra's: We can use Dijkstra’s algorithm to find minimum path from single source with edge weights. The main trick to this problem is we have two constraints to consider in our Dijkstra's implementation. The priority queue must sort by stops and then by cost. We must sort by stops before cost. The issue arises when we find a less cost, but greater stop number path to an intermediary node. From `node0` -> `node1`, we can take `path 0->1`, with 5 cost and 1 stop, or we could take `0->3->1`, with 4 cost and 2 stops. If we update our Dijkstra's shortest path cache to be 4 cost, then we will miss out on a better cost because k = 2 already by taking the shortest path to 1. By taking the shortest cost path to 1 (0->3->1), we already used our 2 stops, so our next node must be our goal of `node2` to not violate `k` constraint. The actual shortest path is `0->1->4->2` where we take a higher cost path to 1, `0->1` cost 5, but `k = 1`.
• Adjacency List: We can use an adjacency list to store the graph, especially when the graph is sparse.
• Adjacency Matrix: We can use an adjacency matrix to store the graph, but a sparse graph will cause an unneeded worst-case runtime.
• Topological Sort: We can use topological sort when a directed graph is used and returns an array of the nodes where each node appears before all the nodes it points to. In order to have a topological sorting, the graph must not contain any cycles.
• Map: We can use a map to store the edges in the graph to lookup equations by name, and store all neighbors by name.
• Union Find: Are there find and union operations here? Can you perform a find operation where you can determine which subset a particular element is in? This can be used for determining if two elements are in the same subset. Can you perform a union operation where you join two subsets into a single subset? Can you check if the two subsets belong to same set? If no, then we cannot perform union.

## 3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General idea: Need to keep track of the number of stops taken to reach a node (city), in addition to the shortest path from the source node

``````To implement Dijkstra, we need a priority queue to pop out the lowest weight node for next search.

In this case, the weight would be the accumulated flight cost.

The node takes a form of `(cost, src, k)`. `cost` is the accumulated cost, `src` is the current node's location, `k` is stop times we left as we only have at most K stops. I also convert `edges` to an adjacent list based graph `g`.

Use a `vis` array to maintain visited nodes to avoid loop. `
vis[x]` record the remaining steps to reach x with the lowest cost.
If `vis[x] >= k`, then no need to visit that case `(start from x with k steps left)` as a better solution has been visited before (more remaining step and lower cost as heappopped beforehand).

And we should initialize `vis[x]` to `0` to ensure visit always stop at a negative `k`.

Once `k` is used up (`k == 0`) or `vis[x] >= k`, we no longer push that node `x` to our queue.
Once a popped cost is our destination, we get our lowest valid cost.``````

⚠️ Common Mistakes

• The current state is being checked against previous state + cost. A common mistake is to check `prev[v]` against `prev[u] + cost(u,v)`. However, in this case we will not get the minimum value of the current state. It's possible that `curr_1 < curr_2 < prev` but, `curr_2` is encountered later than `curr_1`, resulting in it being overwritten.

## 4: I-mplement

Implement the code to solve the algorithm.

``````  class Solution {
int UNVISITED = 10000000;

public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
// build graph, pair first element is denoting destination & second element denotes fare
Map<Integer, List<int[]>> graph = buildGraph(flights, n);

int result = dijkstra(graph, src, dst, k, n);
return result;
}

// apply Dijkstra's
public int dijkstra(
Map<Integer, List<int[]>> graph,
int start, int dest, int k, int n
) {
int[] dist = new int[n];
Arrays.fill(dist, UNVISITED);
dist[start] = 0;

PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> {
if(a[2] == b[2]) return a[1] - b[1]; // if # of stops are equal, sort by cost
return a[2] - b[2]; // otherwise sort first by # of stops
});
pq.offer(new int[]{start, 0, 0});

while(!pq.isEmpty()) {
int[] item = pq.poll();
int node = item[0];
int cost = item[1];
int stops = item[2];

if(stops > k) continue;

List<int[]> neighbors = graph.get(node);

// examine all neighboring edges if possible
for(int[] neighbor : neighbors) {
int neighborNode = neighbor[0];
int neighborCost = neighbor[1];
int totalCost = cost + neighborCost;

// is there a better cost?
if(totalCost < dist[neighborNode]) {
dist[neighborNode] = totalCost;
pq.offer(new int[]{neighborNode, totalCost, stops+1});
}
}
}

return dist[dest] == UNVISITED ? -1 : dist[dest];
}

public Map<Integer, List<int[]>> buildGraph(int[][] flights, int n) {
Map<Integer, List<int[]>> graph = new HashMap<>();

for(int i = 0; i < n; i++) graph.put(i, new ArrayList<>());

for(int[] flight : flights) {
int src = flight[0];
int dest = flight[1];
int cost = flight[2];

}

return graph;
}
}``````
``````import heapq

class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:

# Build the adjacency matrix
adj_matrix = [[0 for _ in range(n)] for _ in range(n)]
for s, d, w in flights:

# Shortest distances array
distances = [float("inf") for _ in range(n)]
current_stops = [float("inf") for _ in range(n)]
distances[src], current_stops[src] = 0, 0

# Data is (cost, stops, node)
minHeap = [(0, 0, src)]

while minHeap:

cost, stops, node = heapq.heappop(minHeap)

# If destination is reached, return the cost to get here
if node == dst:
return cost

# If there are no more steps left, continue
if stops == K + 1:
continue

# Examine and relax all neighboring edges if possible
for nei in range(n):
if adj_matrix[node][nei] > 0:
dU, dV, wUV = cost, distances[nei], adj_matrix[node][nei]

# Better cost?
if dU + wUV < dV:
distances[nei] = dU + wUV
heapq.heappush(minHeap, (dU + wUV, stops + 1, nei))
current_stops[nei] = stops
elif stops < current_stops[nei]:
#  Better steps?
heapq.heappush(minHeap, (dU + wUV, stops + 1, nei))

return -1 if distances[dst] == float("inf") else distances[dst]``````

## 5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors and verify the code works for the happy and edge cases you created in the “Understand” section

## 6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Time Complexity: `O(KE)`, where E is # of edges and K is the # of stops
Space Complexity: `O(V)`, where V is the # of nodes in graph