What is the time complexity of Number of Possible Sets of Closing Branches?

The optimal solution has a time complexity of O(2^n * n^3).

What is the space complexity of Number of Possible Sets of Closing Branches?

The space complexity is O(n^2).

What data structures are used to solve Number of Possible Sets of Closing Branches?

This problem uses Bit Manipulation, Graph, Heap (Priority Queue), Enumeration, Shortest Path. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Possible Sets of Closing Branches?

Insight 1: Bit manipulation is effective for generating all possible subsets of branches (2^n subsets). Insight 2: The Floyd-Warshall algorithm is appropriate for finding all-pairs shortest paths within the active nodes.

Which companies ask LeetCode 2959. Number of Possible Sets of Closing Branches in interviews?

This problem is frequently asked at Atlassian.

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Number of Possible Sets of Closing Branches - LeetCode 2959 Solution

Number of Possible Sets of Closing Branches - Complete Solution Guide

Number of Possible Sets of Closing Branches is LeetCode problem 2959, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a company with n branches across the country, some of which are connected by roads. Initially, all branches are reachable from each other by traveling some roads. The company has realized that they are spending an excessive amount of time traveling between their branches. As a result, they have decided to close down some of these branches ( possibly none ). However, they want to ensure that the remaining branches have a distance of at most maxDistance from each other. The distance betwe

Detailed Explanation

The problem asks us to find the number of subsets of branches in a company such that after closing the branches in the subset, the remaining (active) branches have a pairwise distance of at most `maxDistance`. The input consists of the number of branches `n`, the maximum allowed distance `maxDistance`, and a list of roads connecting the branches with their respective lengths. We need to iterate through all possible subsets of branches, determine the active branches for each subset, calculate the shortest distances between the active branches, and count the subsets for which the pairwise distances are within the `maxDistance` limit. An empty set of active branches (where all branches are closed) is also considered valid.

Solution Approach

The provided code iterates through all possible subsets of branches using bit manipulation. For each subset, it first identifies the active branches. Then, it initializes a distance matrix with the given road lengths and applies the Floyd-Warshall algorithm to compute the shortest distances between all pairs of active branches. Finally, it checks if the distances between all pairs of active branches are within the `maxDistance` limit. If they are, the subset is considered valid, and the count is incremented.

Step-by-Step Algorithm

Step 1: Initialize an adjacency matrix `adj` to represent the graph. Fill it with the edge weights from the input `roads`. Set the diagonal elements to 0 (distance from a node to itself). Set all other pairs to infinity, indicating no direct path.
Step 2: Iterate through all possible subsets of branches using a bitmask. Each bit in the mask represents whether a branch is active (1) or closed (0).
Step 3: For each subset, identify the active branches based on the bitmask. Store the indices of the active branches in a list.
Step 4: Create a distance matrix `dist` and copy the `adj` matrix into it. This matrix will be used for the Floyd-Warshall algorithm.
Step 5: Apply the Floyd-Warshall algorithm to the `dist` matrix, considering only the active branches as intermediate nodes. This will compute the shortest distances between all pairs of active branches.
Step 6: Check if all pairwise distances between active branches are within the `maxDistance` limit. If any distance exceeds `maxDistance`, the subset is invalid.
Step 7: If all pairwise distances are within the limit, increment the count of valid subsets.
Step 8: Return the final count of valid subsets.

Key Insights

Insight 1: Bit manipulation is effective for generating all possible subsets of branches (2^n subsets).
Insight 2: The Floyd-Warshall algorithm is appropriate for finding all-pairs shortest paths within the active nodes.
Insight 3: Handling disconnected graphs or `Integer.MAX_VALUE` (or similar) is crucial to avoid incorrect distance calculations.

Complexity Analysis

Time Complexity: O(2^n * n^3)

Space Complexity: O(n^2)

Topics

This problem involves: Bit Manipulation, Graph, Heap (Priority Queue), Enumeration, Shortest Path.

Companies

Asked at: Atlassian.

Number of Possible Sets of Closing Branches | LeetCode 2959

Hard

Bit Manipulation Graph Heap (Priority Queue)Enumeration Shortest Path

Time: O(2^n * n^3)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int numberOfSets(int n, int maxDistance, int[][] roads) {
        int[][] adj = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                adj[i][j] = Integer.MAX_VALUE;
            }
            adj[i][i] = 0;
        }
        for (int[] road : roads) {
            int u = road[0];
            int v = road[1];
            int w = road[2];
            adj[u][v] = Math.min(adj[u][v], w);
            adj[v][u] = Math.min(adj[v][u], w);
        }

        int count = 0;
        for (int i = 0; i < (1 << n); i++) {
            int[] activeNodes = new int[n];
            int activeCount = 0;
            for (int j = 0; j < n; j++) {
                if ((i >> j) % 2 == 1) {
                    activeNodes[activeCount++] = j;
                }
            }
            
            int[][] dist = new int[n][n];
            for(int u = 0; u < n; u++){
                for(int v = 0; v < n; v++){
                    dist[u][v] = adj[u][v];
                }
            }

            for (int kIndex = 0; kIndex < activeCount; kIndex++) {
                int k = activeNodes[kIndex];
                for (int uIndex = 0; uIndex < activeCount; uIndex++) {
                    int u = activeNodes[uIndex];
                    for (int vIndex = 0; vIndex < activeCount; vIndex++) {
                        int v = activeNodes[vIndex];
                        if (dist[u][k] != Integer.MAX_VALUE && dist[k][v] != Integer.MAX_VALUE) {
                             dist[u][v] = Math.min(dist[u][v], dist[u][k] + dist[k][v]);
                        }
                       
                    }
                }
            }

            boolean isValid = true;
            for (int uIndex = 0; uIndex < activeCount; uIndex++) {
                int u = activeNodes[uIndex];
                for (int vIndex = 0; vIndex < activeCount; vIndex++) {
                    int v = activeNodes[vIndex];
                    if (u != v && dist[u][v] > maxDistance) {
                        isValid = false;
                        break;
                    }
                    if (u == v && dist[u][v] != 0){
                        isValid = false;
                        break;
                    }
                }
                if (!isValid) {
                    break;
                }
            }

            if (isValid) {
                count++;
            }
        }

        return count;
    }
}

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Problem Info

Description

Problem 2959: Number of Possible Sets of Closing Branches - Detailed explanation and solution approaches.

Topics

Bit Manipulation Graph Heap (Priority Queue)Enumeration Shortest Path

Companies

Atlassian

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2959

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

8 steps

1
Step 1: Initialize an adjacency matrix `adj` to represent the graph. Fill it with the edge weights from the input `roads`. Set the diagonal elements to 0 (distance from a node to itself). Set all other pairs to infinity, indicating no direct path.
2
Step 2: Iterate through all possible subsets of branches using a bitmask. Each bit in the mask represents whether a branch is active (1) or closed (0).
3
Step 3: For each subset, identify the active branches based on the bitmask. Store the indices of the active branches in a list.
4
Step 4: Create a distance matrix `dist` and copy the `adj` matrix into it. This matrix will be used for the Floyd-Warshall algorithm.
5
Step 5: Apply the Floyd-Warshall algorithm to the `dist` matrix, considering only the active branches as intermediate nodes. This will compute the shortest distances between all pairs of active branches.
6
Step 6: Check if all pairwise distances between active branches are within the `maxDistance` limit. If any distance exceeds `maxDistance`, the subset is invalid.
7
Step 7: If all pairwise distances are within the limit, increment the count of valid subsets.
8
Step 8: Return the final count of valid subsets.

💡

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(2^n * n^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is dominated by the nested loops for subset generation and the Floyd-Warshall algorithm. The space complexity is determined by the size of the adjacency and distance matrices.

📖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

Dijkstra's Algorithm

Instead of using Floyd-Warshall to find all-pairs shortest paths, we could use Dijkstra's algorithm to find the shortest paths from each active node to all other active nodes. Then, for each pair of active nodes, check if the distance is within maxDistance.

Time: O(2^n * n * (n + m) * log n) where n is the number of nodes and m is the number of edges. This assumes a heap based priority queue implementation of Dijkstra. In our problem, m is at most n^2, but it's usually much smaller. The 2^n comes from iterating over all subsets.Space: O(n^2)

Trade-offs: Dijkstra's algorithm can be more efficient than Floyd-Warshall if the graph is sparse (m << n^2). However, in the worst case (dense graph, m ≈ n^2), Dijkstra's becomes O(n^3 log n), which is worse than Floyd-Warshall's O(n^3). Additionally, implementing Dijkstra requires more code than Floyd-Warshall.

✨

Key Insights

1Insight 1: Bit manipulation is effective for generating all possible subsets of branches (2^n subsets).
2Insight 2: The Floyd-Warshall algorithm is appropriate for finding all-pairs shortest paths within the active nodes.
3Insight 3: Handling disconnected graphs or `Integer.MAX_VALUE` (or similar) is crucial to avoid incorrect distance calculations.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: `n = 1`, `roads = []`. The algorithm correctly handles this by considering both the case where the single branch is active (distance 0 to itself, which is within the limit) and the case where it's closed (no active branches, also valid).
Edge case 2: `maxDistance = 0` and roads exist. This can lead to situations where only very restrictive subsets are allowed. The algorithm should accurately identify and count these subsets.
Edge case 3: Disconnected graphs after closing certain branches. The Floyd-Warshall algorithm with the appropriate handling of 'infinity' ensures that disconnected nodes have a distance greater than `maxDistance`, so invalid subsets are correctly filtered out.
Edge case 4: Empty `roads` array. The initial adjacency matrix will have infinite distances except for the diagonal which is 0. The Floyd-Warshall algorithm will not change this, so only subsets where the distance between each active node is less than or equal to `maxDistance` will be counted.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect initialization of the adjacency matrix. Failing to set the initial distances to infinity (or a sufficiently large value) can lead to incorrect shortest path calculations.

✗Common mistake 2: Not handling disconnected components properly. When closing branches, the graph might become disconnected. The algorithm must ensure that disconnected nodes have a distance greater than `maxDistance` to be considered invalid.

✗Common mistake 3: Not accounting for self-loops (distance from a node to itself). While the problem statement doesn't explicitly mention self-loops, setting `adj[i][i] = 0` is crucial for the Floyd-Warshall algorithm to work correctly.

✗Common mistake 4: Integer overflow. When adding distances in the Floyd-Warshall algorithm, the sum could exceed the maximum integer value. Using `long long` for the distance matrix can help to mitigate this issue, especially for C/C++.

✗Common mistake 5: Not properly filtering for the active nodes when using the intermediate node `k` in the Floyd-Warshall algorithm. The `k`, `u`, and `v` variables should only iterate through the active nodes and not all the nodes from 0 to `n-1`.

✅

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 design. Determining which servers to keep active in a distributed system while ensuring that the latency between any two active servers is below a certain threshold.

🏦Real-world scenario 2: Facility location. Deciding which warehouses or distribution centers to close while maintaining a maximum delivery distance to all remaining customers.

🎮Real-world scenario 3: Resource allocation. Selecting a subset of resources to allocate while ensuring that the communication cost or distance between any two allocated resources is within a specified limit.

💡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

XOR is your friend for "find the single number" problems. Remember: n & (n-1) clears the lowest set bit.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: August 19, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use BFS vs DFS for graph problems?

Use BFS for shortest path in unweighted graphs, level-order processing, or when exploring neighbors systematically. Use DFS for exhaustive search, detecting cycles, topological sort, or when you need to explore complete paths. DFS uses less memory for very wide graphs.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Number of Possible Sets of Closing Branches"?

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

🤖 Need help with this problem?