https://github.com/emahtab/cheapest-flights-within-k-stops
Cheapest Flight with k stops
https://github.com/emahtab/cheapest-flights-within-k-stops
dijkstra-algorithm graph leetcode
Last synced: 5 months ago
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Cheapest Flight with k stops
- Host: GitHub
- URL: https://github.com/emahtab/cheapest-flights-within-k-stops
- Owner: eMahtab
- Created: 2021-09-18T07:00:38.000Z (about 4 years ago)
- Default Branch: main
- Last Pushed: 2024-10-18T17:02:35.000Z (about 1 year ago)
- Last Synced: 2025-02-02T03:26:14.751Z (10 months ago)
- Topics: dijkstra-algorithm, graph, leetcode
- Homepage:
- Size: 2.12 MB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Cheapest flights within k stops
## https://leetcode.com/problems/cheapest-flights-within-k-stops
There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pricei.
You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.
For the above example the cheapest flight from V0 to V2 with at most 2 stops will be (V0 -> V1 -> V4 -> V2 = 7)
```
Constraints:
1. 1 <= n <= 100
2. 0 <= flights.length <= (n * (n - 1) / 2)
3. flights[i].length == 3
4. 0 <= fromi, toi < n
5. fromi != toi
6. 1 <= pricei <= 104
7. There will not be any multiple flights between two cities.
8. 0 <= src, dst, k < n
9. src != dst
```
## Implementation : Dijkstra with stops addition, Runtime Complexity = O(N^2 ∗ Log(N))
```java
class Solution {
public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {
int[][] adjMatrix = new int[n][n];
for (int[] flight : flights) {
adjMatrix[flight[0]][flight[1]] = flight[2];
}
int[] distances = new int[n];
int[] currentStops = new int[n];
Arrays.fill(distances, Integer.MAX_VALUE);
Arrays.fill(currentStops, Integer.MAX_VALUE);
distances[src] = 0;
currentStops[src] = 0;
// The priority queue will contain (node, cost, stops)
PriorityQueue minHeap = new PriorityQueue<>((a, b) -> a[1] - b[1]);
minHeap.offer(new int[] { src, 0, 0 });
while (!minHeap.isEmpty()) {
int[] info = minHeap.poll();
int node = info[0], cost = info[1], stops = info[2];
if (node == dst) {
return cost;
}
if (stops > K) {
continue;
}
for (int nei = 0; nei < n; nei++) {
if (adjMatrix[node][nei] > 0) {
int newCost = cost + adjMatrix[node][nei];
// Relaxation based on cost and stops
if (newCost < distances[nei] || stops + 1 < currentStops[nei]) {
minHeap.offer(new int[] { nei, newCost, stops + 1 });
distances[nei] = newCost;
currentStops[nei] = stops + 1;
}
}
}
}
return -1;
}
}
```
### Code execution
## References:
1. https://leetcode.com/problems/cheapest-flights-within-k-stops/solution
2. https://leetcode.libaoj.in/cheapest-flights-within-k-stops.html