Problem Statement
Alice and Bob share an undirected graph which contains a mix of edges categorized into three types based on their traversal permissions:
- Type 1 edges can only be traversed by Alice.
- Type 2 edges are accessible only to Bob.
- Type 3 edges can be utilized by both Alice and Bob.
This graph is defined using an array called edges, where each element, represented as [typei, ui, vi], denotes an edge of type typei that connects nodes ui and vi. The goal is to find the maximum number of edges that can be removed from this graph while ensuring that the graph still remains traversable by both Alice and Bob, from any node to any other node. If it's impossible for both Alice and Bob to traverse the entire graph, even without removing any edges, the result should be -1.
Examples
Example 1
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.
Example 2
Input:
n = 4, edges = [[3,1,2],[3,2,3],[1,1,4],[2,1,4]]
Output:
0
Explanation:
Notice that removing any edge will not make the graph fully traversable by Alice and Bob.
Example 3
Input:
n = 4, edges = [[3,2,3],[1,1,2],[2,3,4]]
Output:
-1
Explanation:
In the current graph, Alice cannot reach node 4 from the other nodes. Likewise, Bob cannot reach 1. Therefore it's impossible to make the graph fully traversable.
Constraints
1 <= n <= 1051 <= edges.length <= min(105, 3 * n * (n - 1) / 2)edges[i].length == 31 <= typei <= 31 <= ui < vi <= n- All tuples
(typei, ui, vi)are distinct.
Approach and Intuition
To solve this problem, we need an understanding of graph theory, particularly concepts of connectivity and spanning trees. Here’s a step-by-step breakdown of the approach:
Initial Validation: Check if the graph can be fully traversed by Alice and Bob without any removals. If not, return
-1.Edge Priority: Prioritize the use of type 3 edges (usable by both) to maximize the common paths they share in the graph, as this will allow more flexibility in removing type 1 and type 2 edges later on.
Utilizing Union-Find or DFS: A union-find data structure or depth-first search (DFS) can be employed to determine the connected components in the graph:
- For Alice-only and Bob-only graphs constructed using type 1 and type 2 edges respectively, ensure each is initially a spanning tree, i.e., a minimum structure where all nodes are connected without cycles.
- For the mixed graph using type 3 edges, again ensure it forms a minimum spanning tree.
Edge Reduction: After the minimum spanning structure is maintained, further edges (especially those of type 1 and type 2) that do not contribute to connectivity can be removed.
Counting and Comparison: Throughout these operations, keep a count of removable edges. After achieving the spanning structures while integrating as many type 3 edges as possible, compare the reduction count to the maximum number possible while maintaining full traversal capabilities.
The union-find approach is particularly efficient here because it allows for quick union and find operations that help in seamlessly determining and merging connected components, essential for analyzing and adjusting the graph's connectivity dynamically as edges are considered for removal.
Given these matrices, we iteratively adjust the graph, ensuring after each removal step, both Alice and Bob can still traverse the entire graph. If at any step removing more edges violates this requirement, we halt and take the current count as the maximum number of removable edges without breaking traversability.
Solutions
- C++
- Java
class DisjointSet {
vector<int> parent;
vector<int> rank;
int count;
public:
DisjointSet(int size) {
count = size;
for (int i = 0; i <= size; i++) {
parent.push_back(i);
rank.push_back(1);
}
}
int find(int node) {
if (parent[node] == node) {
return node;
}
return parent[node] = find(parent[node]);
}
int unionSets(int node1, int node2) {
int root1 = find(node1);
int root2 = find(node2);
if (root1 == root2) {
return 0;
}
if (rank[root1] > rank[root2]) {
rank[root1] += rank[root2];
parent[root2] = root1;
} else {
rank[root2] += rank[root1];
parent[root1] = root2;
}
count--;
return 1;
}
bool allConnected() {
return count == 1;
}
};
class Solution {
public:
int maximumEdgesToRemove(int n, vector<vector<int>>& edges) {
DisjointSet Alice(n), Bob(n);
int includedEdges = 0;
for (vector<int>& edge : edges) {
if (edge[0] == 3) {
includedEdges += (Alice.unionSets(edge[1], edge[2]) | Bob.unionSets(edge[1], edge[2]));
}
}
for (vector<int>& edge : edges) {
if (edge[0] == 1) {
includedEdges += Alice.unionSets(edge[1], edge[2]);
}
else if (edge[0] == 2) {
includedEdges += Bob.unionSets(edge[1], edge[2]);
}
}
if (Alice.allConnected() && Bob.allConnected()) {
return edges.size() - includedEdges;
}
return -1;
}
};
To solve the problem of removing the maximum number of edges from a graph while ensuring it remains fully traversable by two entities (Alice and Bob), the provided C++ solution utilizes the Disjoint Set (Union-Find) data structure. This structure aids in managing and merging different components of a graph. The solution involves the following key elements:
DisjointSet Class: This internal class handles the operations of union and find. It keeps track of the parent of each node and the rank (to keep the tree flat), crucial for efficient union operations and finding the root of each node.
Major Methods inside DisjointSet:
find: Helps find the root parent of a node, ensuring path compression for efficiency.unionSets: Connects nodes if they belong to different sets. It returns 1 if the nodes were successfully united, implying an edge is truly needed to connect components, otherwise returns 0.allConnected: Checks if all nodes are basically in one set, meaning fully connected.
Solution Class: The main functionality to handle the graph edges and determine how many can be removed while keeping the graph traversable for Alice and Bob.
- A separated
DisjointSetinstance is created for both Alice and Bob to track their distinct sets of connections. - The solution iterates over the edges twice:
- First Iteration: Process the shared edges (type 3), which both Alice and Bob can use. Updates both Alice's and Bob's disjoint sets.
- Second Iteration: Process individual edges for Alice (type 1) and Bob (type 2) separately.
- After processing all edges, the solution checks if both Alice's and Bob's graphs are fully connected by invoking
allConnected()on their respective Disjoint Sets.
- A separated
Outcome:
- If both graphs are fully traversable, computes the number of edges included and subtracts from the total number of edges to get the number of removable edges.
- If it isn’t possible to maintain full traversal capability for both, it returns -1.
This approach efficiently determines the removable edges using the Disjoint Set for union and find operations, ensuring each addition of an edge is necessary for keeping the graph coherent for both parties. Make sure to drive each operation thoughtfully to understand the graph's connectivity dynamics fully. The solution guarantees optimal edge removal while maintaining graph traversal requirements for both Alice and Bob.
class Solution {
public int maxEdgesToRemove(int nodeCount, int[][] edgeInfo) {
UnionController AliceUnion = new UnionController(nodeCount);
UnionController BobUnion = new UnionController(nodeCount);
int essentialEdges = 0;
// Type 3 edges (shared by both Alice and Bob).
for (int[] edge : edgeInfo) {
if (edge[0] == 3) {
essentialEdges += (AliceUnion.makeUnion(edge[1], edge[2]) | BobUnion.makeUnion(edge[1], edge[2]));
}
}
// Type 1 for Alice and Type 2 for Bob.
for (int[] edge : edgeInfo) {
if (edge[0] == 1) {
essentialEdges += AliceUnion.makeUnion(edge[1], edge[2]);
} else if (edge[0] == 2) {
essentialEdges += BobUnion.makeUnion(edge[1], edge[2]);
}
}
// Check if all nodes are connected in Alice's and Bob's graphs
if (AliceUnion.isFullyConnected() && BobUnion.isFullyConnected()) {
return edgeInfo.length - essentialEdges;
}
return -1; // It's not possible to connect all nodes.
}
private class UnionController {
int[] root;
int[] size;
int count; // Number of independent components.
public UnionController(int size) {
count = size;
root = new int[size + 1];
this.size = new int[size + 1];
for (int i = 0; i <= size; i++) {
root[i] = i;
this.size[i] = 1;
}
}
// Find root with path compression
int findRoot(int node) {
if (root[node] != node) {
root[node] = findRoot(root[node]); // Path Compression
}
return root[node];
}
// Union by size, returns 1 if a union was made, otherwise 0.
int makeUnion(int node1, int node2) {
int root1 = findRoot(node1);
int root2 = findRoot(node2);
if (root1 == root2) {
return 0;
}
if (this.size[root1] > this.size[root2]) {
this.size[root1] += this.size[root2];
root[root2] = root1;
} else {
this.size[root2] += this.size[root1];
root[root1] = root2;
}
count--;
return 1;
}
// Check if there is only one component.
boolean isFullyConnected() {
return count == 1;
}
}
}
This Java solution implements a technique to determine the maximum number of edges that can be removed from a graph while ensuring that it remains fully traversable by two separate entities, Alice and Bob. It employs the Union-Find data structure to keep track of connectivity between nodes through union operations and path compression.
* The function `maxEdgesToRemove` initiates two separate Union-Find structures `AliceUnion` and `BobUnion`, one for each user.
* It first iterates over the edges that both Alice and Bob can traverse (Type 3), aggregating these edges if they connect previously unconnected components in either union structure.
* Next, the process handles edges specific to Alice (Type 1) and Bob (Type 2) independently, only adding edges to their respective structures if they connect unconnected components.
* After processing all edges, the solution checks if both graphs are fully connected. If they are, it calculates the maximum number of removable edges by subtracting the count of essential edges (those that caused a union of components) from the total number of edges.
* If not all nodes are connected for either Alice or Bob, the function returns -1, indicating that it's impossible to keep the graph fully traversable without some edges.This method ensures optimal performance and correctness by minimizing unnecessary operations and by leveraging efficient path compression and union by size within the union operations. The solution efficiently determines the feasibility of keeping the graph fully connected with minimal required connections, and then calculates how many surplus edges exist which can be removed.
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