Leetcode #2959: Number of Possible Sets of Closing Branches
In this guide, we solve Leetcode #2959 Number of Possible Sets of Closing Branches in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Graph, Enumeration, Shortest Path, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input: n = 3, maxDistance = 5, roads = [[0,1,2],[1,2,10],[0,2,10]]
Output: 5
Explanation: The possible sets of closing branches are:
- The set [2], after closing, active branches are [0,1] and they are reachable to each other within distance 2.
- The set [0,1], after closing, the active branch is [2].
- The set [1,2], after closing, the active branch is [0].
- The set [0,2], after closing, the active branch is [1].
- The set [0,1,2], after closing, there are no active branches.
It can be proven, that there are only 5 possible sets of closing branches.
Python Solution
class Solution:
def numberOfSets(self, n: int, maxDistance: int, roads: List[List[int]]) -> int:
ans = 0
for mask in range(1 << n):
g = [[inf] * n for _ in range(n)]
for u, v, w in roads:
if mask >> u & 1 and mask >> v & 1:
g[u][v] = min(g[u][v], w)
g[v][u] = min(g[v][u], w)
for k in range(n):
if mask >> k & 1:
g[k][k] = 0
for i in range(n):
for j in range(n):
# g[i][j] = min(g[i][j], g[i][k] + g[k][j])
if g[i][k] + g[k][j] < g[i][j]:
g[i][j] = g[i][k] + g[k][j]
if all(
g[i][j] <= maxDistance
for i in range(n)
for j in range(n)
if mask >> i & 1 and mask >> j & 1
):
ans += 1
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.