How do you solve Count Subtrees With Max Distance Between Cities on LeetCode?

Count Subtrees With Max Distance Between Cities (LeetCode #1617) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree properties can significantly reduce complexity.

What is the brute force solution for Count Subtrees With Max Distance Between Cities?

The brute force approach for Count Subtrees With Max Distance Between Cities is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsets of cities using bitmasking and check if they form valid subtrees. This approach is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Subtrees With Max Distance Between Cities?

Always clarify the problem and constraints before jumping into coding. Discuss your thought process and approach with the interviewer. Test your code with edge cases and explain your reasoning.

What are common mistakes when solving Count Subtrees With Max Distance Between Cities?

Failing to account for connectivity when generating subsets. Not using efficient data structures for graph representation.

#1617

Count Subtrees With Max Distance Between Cities

Hard

Dynamic ProgrammingBit ManipulationTreeEnumerationBitmaskDepth-First SearchDynamic ProgrammingGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

We can use a more efficient approach by leveraging the properties of trees. By using dynamic programming and depth-first search, we can compute the maximum distances in a single traversal.

⚙️

Algorithm

3 steps

1Step 1: Build the adjacency list from the edges.
2Step 2: Use DFS to calculate the maximum distance for each subtree rooted at each node.
3Step 3: Maintain a count of subtrees for each maximum distance found during the DFS.

solution.py28 lines

1# Full working Python code
2from collections import defaultdict
3
4def countSubtrees(n, edges):
5    graph = defaultdict(list)
6    for u, v in edges:
7        graph[u].append(v)
8        graph[v].append(u)
9
10    result = [0] * (n - 1)
11    dp = [0] * (n + 1)
12
13    def dfs(node, parent):
14        max_dist = 0
15        for neighbor in graph[node]:
16            if neighbor == parent:
17                continue
18            child_dist = dfs(neighbor, node) + 1
19            result[child_dist - 1] += 1
20            max_dist = max(max_dist, child_dist)
21        return max_dist
22
23    for i in range(1, n + 1):
24        if dp[i] == 0:
25            dfs(i, -1)
26
27    return result
28

ℹ

Complexity note: The time complexity is O(n) because we traverse each node once during the DFS. The space complexity is O(n) due to the graph representation and the recursion stack.

1Understanding tree properties can significantly reduce complexity.
2Dynamic programming can optimize recursive solutions.

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