How do you solve Number of Ways to Arrive at Destination on LeetCode?

Number of Ways to Arrive at Destination (LeetCode #1976) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using Dijkstra's algorithm helps efficiently find the shortest paths in a weighted graph.

What is the brute force solution for Number of Ways to Arrive at Destination?

The brute force approach for Number of Ways to Arrive at Destination is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible paths from the starting intersection to the destination. We can use depth-first search (DFS) to traverse the graph and count all paths that yield the shortest time, but this method is inefficient for larger graphs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Number of Ways to Arrive at Destination?

Number of Ways to Arrive at Destination uses the following concepts: Dynamic Programming, Graph Theory, Topological Sort, Shortest Path. The recommended patterns to study are: Graph Traversal, Dijkstra's Algorithm, Dynamic Programming.

What are the interview tips for Number of Ways to Arrive at Destination?

Always clarify the problem and constraints before diving into the solution. Discuss your thought process and approach with the interviewer; they may guide you. Practice coding the solution on paper or a whiteboard to simulate real interview conditions.

What are common mistakes when solving Number of Ways to Arrive at Destination?

Not considering all edges that contribute to the shortest path when counting ways. Failing to handle large numbers correctly, leading to overflow errors.

#1976

Number of Ways to Arrive at Destination

Medium

Dynamic ProgrammingGraph TheoryTopological SortShortest PathGraph TraversalDijkstra's AlgorithmDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses Dijkstra's algorithm to find the shortest path from intersection 0 to all other intersections. Then, we can use dynamic programming to count the number of ways to reach the destination using only the edges that contribute to the shortest path.

⚙️

Algorithm

3 steps

1Step 1: Use Dijkstra's algorithm to find the shortest time from intersection 0 to all intersections.
2Step 2: Create a DP array where dp[i] represents the number of ways to reach intersection i in the shortest time.
3Step 3: Traverse the graph again, and for each edge that contributes to the shortest path, update the DP array accordingly.

solution.py32 lines

1# Full working Python code
2import heapq
3from collections import defaultdict
4
5def countPaths(n, roads):
6    graph = defaultdict(list)
7    for u, v, time in roads:
8        graph[u].append((v, time))
9        graph[v].append((u, time))
10
11    # Step 1: Dijkstra's algorithm
12    dist = [float('inf')] * n
13    dist[0] = 0
14    pq = [(0, 0)]  # (time, node)
15    while pq:
16        time, node = heapq.heappop(pq)
17        for neighbor, travel_time in graph[node]:
18            if time + travel_time < dist[neighbor]:
19                dist[neighbor] = time + travel_time
20                heapq.heappush(pq, (dist[neighbor], neighbor))
21
22    # Step 2: DP array
23    dp = [0] * n
24    dp[0] = 1
25
26    # Step 3: Count paths
27    for node in range(n):
28        for neighbor, travel_time in graph[node]:
29            if dist[node] + travel_time == dist[neighbor]:
30                dp[neighbor] = (dp[neighbor] + dp[node]) % (10**9 + 7)
31
32    return dp[n - 1]

ℹ

Complexity note: The time complexity is O(n log n) due to Dijkstra's algorithm, which uses a priority queue to efficiently find the shortest paths. The space complexity is O(n) for storing the graph and the DP array.

1Using Dijkstra's algorithm helps efficiently find the shortest paths in a weighted graph.
2Dynamic programming allows us to count the number of ways to reach the destination using only valid edges.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.