How do you solve Minimum Score of a Path Between Two Cities on LeetCode?

Minimum Score of a Path Between Two Cities (LeetCode #2492) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: The minimum score is determined by the weakest link in the path, which is the minimum distance road.

What is the time complexity of Minimum Score of a Path Between Two Cities?

The optimal solution for Minimum Score of a Path Between Two Cities runs in O(n + m) time and uses O(n) space. The time complexity is O(n + m) where n is the number of cities and m is the number of roads, as we traverse each node and edge once during BFS.

What is the brute force solution for Minimum Score of a Path Between Two Cities?

The brute force approach for Minimum Score of a Path Between Two Cities is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths between the two cities and calculating the minimum road distance for each path. This can be visualized as trying every possible route to find the best one. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Minimum Score of a Path Between Two Cities?

Minimum Score of a Path Between Two Cities uses the following concepts: Depth-First Search, Breadth-First Search, Union-Find, Graph Theory. The recommended patterns to study are: Graph Traversal, Breadth-First Search, Depth-First Search.

What are the interview tips for Minimum Score of a Path Between Two Cities?

Clearly explain your thought process and the reasoning behind your chosen approach. Be prepared to discuss the trade-offs between different algorithms. Practice implementing graph traversal algorithms like BFS and DFS.

What are common mistakes when solving Minimum Score of a Path Between Two Cities?

Not considering all possible paths can lead to incorrect minimum scores. Failing to properly manage visited nodes can cause infinite loops.

#2492

Minimum Score of a Path Between Two Cities

Medium

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryGraph TraversalBreadth-First SearchDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n + m)Space O(n)

The optimal solution leverages a graph traversal technique (like BFS) to find the minimum score efficiently. Instead of exploring all paths, we only need to find the connected component between cities 1 and n and track the minimum edge weight.

⚙️

Algorithm

3 steps

1Step 1: Build the graph using an adjacency list from the roads data.
2Step 2: Use Breadth-First Search (BFS) starting from city 1 to explore all reachable cities.
3Step 3: During the BFS, track the minimum distance of the roads used to reach city n.

solution.py20 lines

1def minScore(n, roads):
2    from collections import defaultdict, deque
3    graph = defaultdict(list)
4    for a, b, d in roads:
5        graph[a].append((b, d))
6        graph[b].append((a, d))
7
8    min_distance = float('inf')
9    queue = deque([1])
10    visited = set([1])
11
12    while queue:
13        city = queue.popleft()
14        for neighbor, distance in graph[city]:
15            min_distance = min(min_distance, distance)
16            if neighbor not in visited:
17                visited.add(neighbor)
18                queue.append(neighbor)
19
20    return min_distance

ℹ

Complexity note: The time complexity is O(n + m) where n is the number of cities and m is the number of roads, as we traverse each node and edge once during BFS.

1The minimum score is determined by the weakest link in the path, which is the minimum distance road.
2Using BFS allows us to efficiently explore the graph without redundant paths.

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