Leetcode #1579: Remove Max Number of Edges to Keep Graph Fully Traversable
In this guide, we solve Leetcode #1579 Remove Max Number of Edges to Keep Graph Fully Traversable 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
Alice and Bob have an undirected graph of n nodes and three types of edges: Type 1: Can be traversed by Alice only. Type 2: Can be traversed by Bob only.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Union Find, Graph
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,3],[1,2,4],[1,1,2],[2,3,4]]
Output: 2
Explanation: If we remove the 2 edges [1,1,2] and [1,1,3]. The graph will still be fully traversable by Alice and Bob. Removing any additional edge will not make it so. So the maximum number of edges we can remove is 2.
Python Solution
class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n
self.cnt = n
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a, b):
pa, pb = self.find(a - 1), self.find(b - 1)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
self.cnt -= 1
return True
class Solution:
def maxNumEdgesToRemove(self, n: int, edges: List[List[int]]) -> int:
ufa = UnionFind(n)
ufb = UnionFind(n)
ans = 0
for t, u, v in edges:
if t == 3:
if ufa.union(u, v):
ufb.union(u, v)
else:
ans += 1
for t, u, v in edges:
if t == 1:
ans += not ufa.union(u, v)
if t == 2:
ans += not ufb.union(u, v)
return ans if ufa.cnt == 1 and ufb.cnt == 1 else -1
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.