Unit 10 Session 1 Advanced (Click for link to problem statements)
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?
flights[i][j] = 1 mean?
i to airport j.-1 if no path exists.HAPPY CASE
Input: flights = [
[0, 1, 0, 0], # Airport 0
[0, 0, 1, 0], # Airport 1
[0, 0, 0, 1], # Airport 2
[0, 0, 0, 0] # Airport 3
]
start = 0
destination = 3
Output: 3
Explanation: The shortest path from airport 0 to airport 3 requires 3 flights (0 -> 1 -> 2 -> 3).
EDGE CASE
Input: flights = [
[0, 0, 0, 0], # Airport 0
[0, 0, 1, 0], # Airport 1
[0, 0, 0, 1], # Airport 2
[0, 0, 0, 0] # Airport 3
]
start = 0
destination = 3
Output: -1
Explanation: There is no valid path from airport 0 to airport 3.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 Shortest Path problems, we want to consider the following approaches:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: We will use BFS to find the shortest path between the starting and destination airports. BFS explores all airports layer by layer, ensuring that we find the shortest path (minimum number of flights).
1) Edge case: If the start airport is the same as the destination, return `0` (no flights needed).
2) Initialize a queue with `(start_airport, num_flights)` and a set `visited` to track the visited airports.
3) Perform BFS:
a) Dequeue the current airport and number of flights.
b) If the current airport is the destination, return the number of flights.
c) For each neighboring airport (next_airport), if there is a direct flight and the neighbor hasn't been visited, add it to the queue and mark it as visited.
4) If BFS completes without finding the destination, return `-1`.⚠️ Common Mistakes
Implement the code to solve the algorithm.
from collections import deque
def counting_flights(flights, start, destination):
n = len(flights)
# Step 1: Edge case - if start and destination are the same, no num_flights needed
if start == destination:
return 0
# Step 2: Initialize BFS structures
queue = deque([(start, 0)]) # (current airport, current number of num_flights)
visited = set([start])
# Step 3: BFS loop
while queue:
current_airport, num_flights = queue.popleft()
# Explore neighbors (direct flights from current airport)
for next_airport in range(n):
if flights[current_airport][next_airport] == 1 and next_airport not in visited:
if next_airport == destination:
return num_flights + 1
queue.append((next_airport, num_flights + 1))
visited.add(next_airport)
# Step 4: If we complete BFS without finding the destination, return -1
return -1Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
flights = [
[0, 1, 0, 0], # Airport 0
[0, 0, 1, 0], # Airport 1
[0, 0, 0, 1], # Airport 2
[0, 0, 0, 0] # Airport 3
]
print(counting_flights(flights, 0, 3)) # Output: 3
flights = [
[0, 0, 0, 0], # Airport 0
[0, 0, 1, 0], # Airport 1
[0, 0, 0, 1], # Airport 2
[0, 0, 0, 0] # Airport 3
]
print(counting_flights(flights, 0, 3)) # Output: -1Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(n^2), where n is the number of airports. BFS explores all possible paths, and for each airport, we check all its neighbors.O(n) for the queue and visited set, as we store up to n airports.