How do you solve Number of Possible Sets of Closing Branches on LeetCode?

Number of Possible Sets of Closing Branches (LeetCode #2959) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³ + 2^n * n²) time complexity and O(n²) space complexity. Key insight: Using bit manipulation can significantly reduce the complexity of checking subsets.

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

The optimal solution for Number of Possible Sets of Closing Branches runs in O(n³ + 2^n * n²) time and uses O(n²) space. Floyd-Warshall runs in O(n³) and checking each subset takes O(2^n * n²) due to the nested loops for pair checking.

What is the brute force solution for Number of Possible Sets of Closing Branches?

The brute force approach for Number of Possible Sets of Closing Branches is Brute Force, with O(n²) time complexity and O(1) space complexity. We will try every possible combination of branches to close and check if the remaining branches are within the max distance. This is a straightforward approach but can be inefficient as it explores all subsets. The optimal Optimal Solution approach improves this to O(n³ + 2^n * n²).

What are the interview tips for Number of Possible Sets of Closing Branches?

Always clarify the problem constraints and edge cases. Explain your thought process while coding to demonstrate understanding. Practice implementing algorithms like Floyd-Warshall to become comfortable with them.

What are common mistakes when solving Number of Possible Sets of Closing Branches?

Failing to consider edge cases where only one or no branches are closed. Not correctly implementing the Floyd-Warshall algorithm.

#2959

Number of Possible Sets of Closing Branches

Hard

Bit ManipulationGraph TheoryHeap (Priority Queue)EnumerationShortest PathBit ManipulationGraph TheoryDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³ + 2^n * n²)
Space	O(1)	O(n²)

💡

Intuition

Time O(n³ + 2^n * n²)Space O(n²)

Instead of checking all subsets, we can use a more efficient approach by leveraging the Floyd-Warshall algorithm to precompute shortest paths and then use bit manipulation to efficiently check subsets of branches.

⚙️

Algorithm

3 steps

1Step 1: Build a graph from the roads and compute the shortest paths between all pairs of branches using Floyd-Warshall.
2Step 2: For each subset of branches, check if all pairs of branches in the subset are within maxDistance using the precomputed shortest paths.
3Step 3: Count valid subsets.

solution.py37 lines

1# Full working Python code
2import itertools
3import sys
4
5def floyd_warshall(n, graph):
6    dist = [[sys.maxsize] * n for _ in range(n)]
7    for u in range(n):
8        dist[u][u] = 0
9    for u, v, w in graph:
10        dist[u][v] = min(dist[u][v], w)
11        dist[v][u] = min(dist[v][u], w)
12    for k in range(n):
13        for i in range(n):
14            for j in range(n):
15                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
16    return dist
17
18def count_possible_sets(n, maxDistance, roads):
19    graph = roads
20    dist = floyd_warshall(n, graph)
21    count = 0
22    for mask in range(1 << n):
23        valid = True
24        for i in range(n):
25            for j in range(i + 1, n):
26                if (mask & (1 << i)) and (mask & (1 << j)):
27                    if dist[i][j] > maxDistance:
28                        valid = False
29                        break
30            if not valid:
31                break
32        if valid:
33            count += 1
34    return count
35
36# Example usage
37print(count_possible_sets(3, 5, [[0, 1, 2], [1, 2, 10], [0, 2, 10]]))

ℹ

Complexity note: Floyd-Warshall runs in O(n³) and checking each subset takes O(2^n * n²) due to the nested loops for pair checking.

1Using bit manipulation can significantly reduce the complexity of checking subsets.
2Precomputing distances allows for quick validation of branch sets.

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