#1617

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Count Subtrees With Max Distance Between Cities

This problem asks you to count subtrees in a tree structure where the maximum distance between any two cities matches specific values.

dynamic programming bit manipulation tree enumeration bitmask

Problem Statement

You are given a tree with n cities, each numbered from 1 to n. The tree is described by an array of edges where each edge connects two cities, forming a bidirectional path. The tree structure ensures there is a unique path between any pair of cities. Your task is to count how many subtrees have a specific maximum distance between any two cities. A subtree consists of any subset of cities that are connected within themselves.

For each distance d from 1 to n-1, determine how many subtrees have the maximum distance between any two cities in the subtree equal to d. The key challenge is to efficiently count these subtrees using dynamic programming and bitmasking techniques while ensuring each subtree's validity by tracking the maximum distance between the cities in the subset.

Examples

Example 1

Input: n = 4, edges = [[1,2],[2,3],[2,4]]

Output: [3,4,0]

The subtrees with subsets {1,2}, {2,3} and {2,4} have a max distance of 1.

The subtrees with subsets {1,2,3}, {1,2,4}, {2,3,4} and {1,2,3,4} have a max distance of 2.

No subtree has two nodes where the max distance between them is 3.

Example 2

Input: n = 2, edges = [[1,2]]

Output: [1]

Example details omitted.

Example 3

Input: n = 3, edges = [[1,2],[2,3]]

Output: [2,1]

Example details omitted.

Constraints

2 <= n <= 15
edges.length == n-1
edges[i].length == 2
1 <= ui, vi <= n
All pairs (ui, vi) are distinct.

Solution Approach

Bitmasking to Iterate Over All Subtrees

Use bitmasking to represent every possible subset of cities. This allows you to efficiently generate all possible subtrees and track which cities are included in each subset. For each bitmask, check if the subset of cities forms a valid subtree and calculate the maximum distance between cities in that subtree.

Dynamic Programming for Subtree Distance Calculation

Implement dynamic programming to calculate the maximum distance between cities in each valid subtree. As you iterate over each possible subtree, store precomputed values to avoid redundant calculations. This improves the efficiency of the algorithm, especially when working with a larger number of cities.

Optimization for Valid Subtree Validation

For each generated subtree, determine if it is valid by checking if all cities in the subset are connected. This can be done using depth-first search (DFS) or breadth-first search (BFS) to ensure the subset of cities forms a connected graph. If valid, calculate the maximum distance between any two cities in the subtree.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of this solution depend on the number of cities n. Using bitmasking, you need to consider all subsets of cities, which takes O(2^n) time. For each subset, you must check the maximum distance between cities, which requires traversing the tree. The space complexity is determined by the number of subtrees and the space needed for dynamic programming states.

What Interviewers Usually Probe

Look for the candidate's understanding of bitmasking and dynamic programming techniques.
Assess the candidate's ability to handle tree traversal efficiently within constraints.
Evaluate their ability to optimize the algorithm, especially considering time and space complexity.

Common Pitfalls or Variants

Common pitfalls

Not efficiently managing the bitmasking process and trying to iterate through all subsets without considering optimizations.
Failing to validate whether a subset forms a connected graph before calculating the maximum distance between cities.
Overlooking the importance of dynamic programming in storing precomputed values to avoid redundant calculations.

Follow-up variants

Increase the number of cities to test the scalability of the solution.
Consider additional constraints such as weighted edges or directed edges to modify the problem.
Modify the problem to count subtrees with a fixed number of cities and a maximum distance limit.

FAQ

What is the pattern for solving 'Count Subtrees With Max Distance Between Cities'?

The pattern revolves around binary-tree traversal and state tracking, specifically using bitmasking and dynamic programming to efficiently find subtrees and their maximum distances.

How do I use bitmasking for this problem?

Bitmasking is used to represent every possible subset of cities, which helps in efficiently iterating over potential subtrees and checking their validity and maximum distance.

How do I determine if a subtree is valid?

To determine if a subtree is valid, check if all cities in the subset are connected. This can be done using DFS or BFS to ensure the subset forms a valid tree structure.

What is the time complexity of this approach?

The time complexity is O(2^n * n), where n is the number of cities. This comes from iterating over all subsets using bitmasking and performing a tree traversal for each subset.

Can this problem be optimized further?

Yes, further optimizations can be achieved by reducing redundant calculations using dynamic programming and efficiently managing the bitmasking process.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def countSubgraphsForEachDiameter(
        self, n: int, edges: List[List[int]]
    ) -> List[int]:
        def dfs(u: int, d: int = 0):
            nonlocal mx, nxt, msk
            if mx < d:
                mx, nxt = d, u
            msk ^= 1 << u
            for v in g[u]:
                if msk >> v & 1:
                    dfs(v, d + 1)

        g = defaultdict(list)
        for u, v in edges:
            u, v = u - 1, v - 1
            g[u].append(v)
            g[v].append(u)
        ans = [0] * (n - 1)
        nxt = mx = 0
        for mask in range(1, 1 << n):
            if mask & (mask - 1) == 0:
                continue
            msk, mx = mask, 0
            cur = msk.bit_length() - 1
            dfs(cur)
            if msk == 0:
                msk, mx = mask, 0
                dfs(nxt)
                ans[mx - 1] += 1
        return ans

Solution 2

#### Python3

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class Solution:
    def countSubgraphsForEachDiameter(
        self, n: int, edges: List[List[int]]
    ) -> List[int]:
        def dfs(u: int, d: int = 0):
            nonlocal mx, nxt, msk
            if mx < d:
                mx, nxt = d, u
            msk ^= 1 << u
            for v in g[u]:
                if msk >> v & 1:
                    dfs(v, d + 1)

        g = defaultdict(list)
        for u, v in edges:
            u, v = u - 1, v - 1
            g[u].append(v)
            g[v].append(u)
        ans = [0] * (n - 1)
        nxt = mx = 0
        for mask in range(1, 1 << n):
            if mask & (mask - 1) == 0:
                continue
            msk, mx = mask, 0
            cur = msk.bit_length() - 1
            dfs(cur)
            if msk == 0:
                msk, mx = mask, 0
                dfs(nxt)
                ans[mx - 1] += 1
        return ans

Continue Practicing

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