What is the time complexity of Count Pairs of Connectable Servers in a Weighted Tree Network?

The optimal solution for Count Pairs of Connectable Servers in a Weighted Tree Network runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse each node once during the DFS. The space complexity is O(n) due to the storage of the graph and subtree counts.

What is the brute force solution for Count Pairs of Connectable Servers in a Weighted Tree Network?

The brute force approach for Count Pairs of Connectable Servers in a Weighted Tree Network is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we check every possible pair of servers for each server to see if they are connectable. This involves a lot of repeated calculations and is inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Pairs of Connectable Servers in a Weighted Tree Network?

Count Pairs of Connectable Servers in a Weighted Tree Network uses the following concepts: Array, Tree, Depth-First Search. The recommended patterns to study are: Depth-First Search, Tree Traversal, Dynamic Programming on Trees.

What are the interview tips for Count Pairs of Connectable Servers in a Weighted Tree Network?

Always clarify the problem requirements and constraints before jumping into coding. Think about edge cases, especially with tree structures, like single child nodes or deep trees. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Count Pairs of Connectable Servers in a Weighted Tree Network?

Failing to account for the conditions of connectability correctly, especially the edge-sharing condition. Not properly implementing the DFS or miscalculating distances.

#3067

Count Pairs of Connectable Servers in a Weighted Tree Network

Medium

ArrayTreeDepth-First SearchDepth-First SearchTree TraversalDynamic Programming on Trees

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)

The optimal approach uses Depth-First Search (DFS) to count the number of nodes in each subtree that are at distances divisible by signalSpeed. This avoids redundant calculations and efficiently counts connectable pairs.

⚙️

Algorithm

4 steps

1Step 1: Build the tree from the edges.
2Step 2: For each server c, run DFS to count nodes in its subtree whose distances from c are divisible by signalSpeed.
3Step 3: For each child of c, calculate the number of connectable pairs using the formula: ((S - num[c_i]) * num[c_i]) / 2, where S is the total count of nodes in the subtree.
4Step 4: Store the result for each server.

solution.py34 lines

1# Full working Python code
2from collections import defaultdict
3
4def countPairsOptimal(edges, signalSpeed):
5    n = len(edges) + 1
6    graph = defaultdict(list)
7    for a, b, weight in edges:
8        graph[a].append((b, weight))
9        graph[b].append((a, weight))
10
11    count = [0] * n
12    subtree_count = [0] * n
13
14    def dfs(node, parent, depth):
15        total = 1
16        for neighbor, weight in graph[node]:
17            if neighbor != parent:
18                child_count = dfs(neighbor, node, depth + weight)
19                if (depth + weight) % signalSpeed == 0:
20                    count[node] += child_count
21                total += child_count
22        subtree_count[node] = total
23        return total
24
25    dfs(0, -1, 0)
26
27    for c in range(n):
28        total_pairs = 0
29        for neighbor, weight in graph[c]:
30            if subtree_count[neighbor] > 0:
31                total_pairs += (subtree_count[c] - subtree_count[neighbor]) * subtree_count[neighbor]
32        count[c] += total_pairs // 2
33
34    return count

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 storage of the graph and subtree counts.

1Understanding the structure of trees is crucial for efficiently solving problems related to them.
2Using DFS allows us to gather necessary information about the tree in a single traversal, reducing redundant calculations.

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