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

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Count Pairs of Connectable Servers in a Weighted Tree Network?

This problem uses Array, Tree, Depth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Pairs of Connectable Servers in a Weighted Tree Network?

Insight 1: The core idea is to iterate through each server 'c' and consider it as the central server. For each central server, we need to explore the paths to other servers. Insight 2: Since the path from 'c' to 'a' and 'c' to 'b' must not share edges, we can treat each neighbor of 'c' as the starting point of a separate subtree. We count how many servers in each subtree have a distance from 'c' divisible by 'signalSpeed'.

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Count Pairs of Connectable Servers in a Weighted Tree Network - LeetCode 3067 Solution

Q: How to solve Count Pairs of Connectable Servers in a Weighted Tree Network?

The provided code uses an iterative Depth-First Search (DFS) approach to solve this problem. For each server 'i', it considers 'i' as the central server and iterates through its neighbors. Each neighbor represents the starting point of a distinct subtree. A DFS is performed starting from each neighbor to compute the distance to all other servers in that subtree. If the distance from 'i' to a server in a subtree is divisible by 'signalSpeed', it increments the count for that subtree. After processing all subtrees, it calculates the number of connectable server pairs by multiplying the number of servers divisible by signalSpeed in each subtree with the prefix sum of counts from previously processed subtrees. This ensures no shared edges between paths.

Q: Which companies ask LeetCode 3067. Count Pairs of Connectable Servers in a Weighted Tree Network in interviews?

This problem is frequently asked at UBS, thoughtspot.

Count Pairs of Connectable Servers in a Weighted Tree Network - Complete Solution Guide

Count Pairs of Connectable Servers in a Weighted Tree Network is LeetCode problem 3067, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an unrooted weighted tree with n vertices representing servers numbered from 0 to n - 1 , an array edges where edges[i] = [a i , b i , weight i ] represents a bidirectional edge between vertices a i and b i of weight weight i . You are also given an integer signalSpeed . Two servers a and b are connectable through a server c if: a < b , a != c and b != c . The distance from c to a is divisible by signalSpeed . The distance from c to b is divisible by signalSpeed . The path from c t

Detailed Explanation

The problem asks us to find the number of 'connectable' server pairs for each server in a weighted, unrooted tree network. Two servers, 'a' and 'b', are connectable through a server 'c' if: 1) a < b, a != c, and b != c; 2) the distance from c to a is divisible by a given 'signalSpeed'; 3) the distance from c to b is also divisible by 'signalSpeed'; and 4) the paths from c to a and c to b don't share any edges. We are given the tree's edges and their weights, along with the 'signalSpeed'. The output should be an array where each element `count[i]` represents the number of server pairs connectable through server 'i'.

Solution Approach

The provided code uses an iterative Depth-First Search (DFS) approach to solve this problem. For each server 'i', it considers 'i' as the central server and iterates through its neighbors. Each neighbor represents the starting point of a distinct subtree. A DFS is performed starting from each neighbor to compute the distance to all other servers in that subtree. If the distance from 'i' to a server in a subtree is divisible by 'signalSpeed', it increments the count for that subtree. After processing all subtrees, it calculates the number of connectable server pairs by multiplying the number of servers divisible by signalSpeed in each subtree with the prefix sum of counts from previously processed subtrees. This ensures no shared edges between paths.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the input 'edges'. The adjacency list stores the neighbors and their corresponding edge weights for each server.
Step 2: Iterate through each server 'i' from 0 to n-1. This server will be considered as the central server 'c'.
Step 3: For each central server 'i', iterate through its neighbors in the adjacency list.
Step 4: For each neighbor, perform an iterative DFS to explore the subtree rooted at that neighbor. Keep track of the distance from server 'i' to each server in the subtree.
Step 5: During the DFS, if the distance from server 'i' to a server 'u' is divisible by 'signalSpeed', increment the 'current_subtree_count' for that subtree.
Step 6: After the DFS completes for a subtree, store the 'current_subtree_count' in a list called 'subtree_counts'.
Step 7: After processing all neighbors (all subtrees) of server 'i', check if there are at least two subtrees (i.e., the size of 'subtree_counts' is greater than or equal to 2). If not, there are no connectable pairs through server 'i'.
Step 8: If there are at least two subtrees, calculate the number of connectable pairs by iterating through the 'subtree_counts' list. For each count, multiply it by the 'prefix_sum' (sum of counts from previously processed subtrees) and add the result to 'total_pairs'. Then, update the 'prefix_sum' by adding the current count.
Step 9: Store the 'total_pairs' in the 'ans' array at index 'i'.
Step 10: After processing all servers, return the 'ans' array.

Key Insights

Insight 1: The core idea is to iterate through each server 'c' and consider it as the central server. For each central server, we need to explore the paths to other servers.
Insight 2: Since the path from 'c' to 'a' and 'c' to 'b' must not share edges, we can treat each neighbor of 'c' as the starting point of a separate subtree. We count how many servers in each subtree have a distance from 'c' divisible by 'signalSpeed'.
Insight 3: The problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS) to calculate the distances from each server 'c' to all other servers in its subtrees. Iterative DFS using a stack avoids potential stack overflow issues with recursive DFS for larger trees.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Tree, Depth-First Search.

Companies

Asked at: UBS, thoughtspot.

Count Pairs of Connectable Servers in a Weighted Tree Network | LeetCode 3067

Medium

Array Tree Depth-First Search

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

class Solution {
    public int[] countPairsOfConnectableServers(int[][] edges, int signalSpeed) {
        int n = edges.length + 1;
        List<List<Pair>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int w = edge[2];
            adj.get(u).add(new Pair(v, w));
            adj.get(v).add(new Pair(u, w));
        }

        int[] ans = new int[n];

        for (int i = 0; i < n; i++) {
            if (adj.get(i).size() < 2) {
                continue;
            }

            List<Integer> subtreeCounts = new ArrayList<>();

            for (Pair neighborPair : adj.get(i)) {
                int neighbor = neighborPair.node;
                int weight = neighborPair.weight;
                int currentSubtreeCount = 0;

                // Iterative DFS using a stack
                // (current_node, parent_node, distance_from_server_i)
                Stack<Triple> stack = new Stack<>();
                stack.push(new Triple(neighbor, i, weight));

                while (!stack.isEmpty()) {
                    Triple triple = stack.pop();
                    int u = triple.node;
                    int p = triple.parent;
                    int dist = triple.distance;

                    if (dist % signalSpeed == 0) {
                        currentSubtreeCount += 1;
                    }

                    for (Pair vPair : adj.get(u)) {
                        int v = vPair.node;
                        int w = vPair.weight;
                        if (v != p) {
                            stack.push(new Triple(v, u, dist + w));
                        }
                    }
                }

                if (currentSubtreeCount > 0) {
                    subtreeCounts.add(currentSubtreeCount);
                }
            }

            if (subtreeCounts.size() < 2) {
                continue;
            }

            int totalPairs = 0;
            int prefixSum = 0;
            for (int count : subtreeCounts) {
                totalPairs += count * prefixSum;
                prefixSum += count;
            }

            ans[i] = totalPairs;
        }

        return ans;
    }

    private static class Pair {
        int node;
        int weight;

        public Pair(int node, int weight) {
            this.node = node;
            this.weight = weight;
        }
    }
    
    private static class Triple {
        int node;
        int parent;
        int distance;
        
        public Triple(int node, int parent, int distance) {
            this.node = node;
            this.parent = parent;
            this.distance = distance;
        }
    }
}

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Problem Info

Description

Problem 3067: Count Pairs of Connectable Servers in a Weighted Tree Network - Detailed explanation and solution approaches.

Topics

Array Tree Depth-First Search

Companies

UBSthoughtspot

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3067

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Build the adjacency list representation of the tree from the input 'edges'. The adjacency list stores the neighbors and their corresponding edge weights for each server.
2
Step 2: Iterate through each server 'i' from 0 to n-1. This server will be considered as the central server 'c'.
3
Step 3: For each central server 'i', iterate through its neighbors in the adjacency list.
4
Step 4: For each neighbor, perform an iterative DFS to explore the subtree rooted at that neighbor. Keep track of the distance from server 'i' to each server in the subtree.
5
Step 5: During the DFS, if the distance from server 'i' to a server 'u' is divisible by 'signalSpeed', increment the 'current_subtree_count' for that subtree.
6
Step 6: After the DFS completes for a subtree, store the 'current_subtree_count' in a list called 'subtree_counts'.
7
Step 7: After processing all neighbors (all subtrees) of server 'i', check if there are at least two subtrees (i.e., the size of 'subtree_counts' is greater than or equal to 2). If not, there are no connectable pairs through server 'i'.
8
Step 8: If there are at least two subtrees, calculate the number of connectable pairs by iterating through the 'subtree_counts' list. For each count, multiply it by the 'prefix_sum' (sum of counts from previously processed subtrees) and add the result to 'total_pairs'. Then, update the 'prefix_sum' by adding the current count.
9
Step 9: Store the 'total_pairs' in the 'ans' array at index 'i'.
10
Step 10: After processing all servers, return the 'ans' array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops, with the outer loop iterating through each server and the inner loop performing DFS. The space complexity is mainly due to the adjacency list, the subtree counts array, and the stack used during DFS.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

Instead of DFS, we can use BFS to traverse the subtrees and calculate the distances. The overall logic remains the same, but we use a queue instead of a stack.

Time: O(n^2)Space: O(n)

Trade-offs: BFS can be considered if you prefer a queue-based approach over a stack. The practical performance difference between iterative DFS and BFS in this case is likely to be negligible.

✨

Key Insights

1Insight 1: The core idea is to iterate through each server 'c' and consider it as the central server. For each central server, we need to explore the paths to other servers.
2Insight 2: Since the path from 'c' to 'a' and 'c' to 'b' must not share edges, we can treat each neighbor of 'c' as the starting point of a separate subtree. We count how many servers in each subtree have a distance from 'c' divisible by 'signalSpeed'.
3Insight 3: The problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS) to calculate the distances from each server 'c' to all other servers in its subtrees. Iterative DFS using a stack avoids potential stack overflow issues with recursive DFS for larger trees.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: A server with a degree less than 2. If a server has fewer than two neighbors, it cannot be a connecting server, so the algorithm correctly skips such servers.
Edge case 2: signalSpeed is very large compared to the edge weights. In this case, very few nodes in the subtrees will have distances divisible by signalSpeed, potentially resulting in subtree_counts being all zeros and leading to a zero value for total_pairs.
Edge case 3: Empty graph (n=1). The given constraints specify n>=2 so this case should not exist. If it did, it would need explicit handling.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using recursive DFS without considering potential stack overflow issues for large trees. Iterative DFS using a stack is a safer approach.

✗Common mistake 2: Incorrectly calculating the number of connectable pairs by not considering the constraint that paths should not share edges. Treating each neighbor as a root of a separate subtree ensures this constraint is met.

✗Common mistake 3: Not handling the case where a server has fewer than two neighbors, which can lead to incorrect results or errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network analysis where you need to find nodes that can facilitate communication between other nodes subject to latency or bandwidth constraints (represented by 'signalSpeed').

🏦Real-world scenario 2: Social network analysis where you want to find influential individuals who can connect other individuals based on specific criteria or limitations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Pairs of Connectable Servers in a Weighted Tree Network"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?