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This approach uses a Union-Find (Disjoint Set Union, DSU) structure to detect cycles and check for nodes with two parents. The goal is to handle two situations: a node having two parents, and a cycle existing in the graph. We iterate through the edges to identify a node with two parents and mark the offending edge. Then, we use the Union-Find structure to track cycles and find the redundant connection based on the identified edges.
Time Complexity: O(n), where n is the number of edges.
Space Complexity: O(n), for storing the parent and rank arrays.
1class UnionFind {
2 private int[] parent;
3 private int[] rank;
4
5 public UnionFind(int size) {
6 parent = new int[size];
7 rank = new int[size];
8 for (int i = 0; i < size; i++) {
9 parent[i] = i;
10 rank[i] = 1;
11 }
12 }
13
14 public int find(int x) {
15 if (parent[x] != x) {
16 parent[x] = find(parent[x]);
17 }
18 return parent[x];
19 }
20
21 public boolean union(int x, int y) {
22 int rootX = find(x);
23 int rootY = find(y);
24 if (rootX == rootY) return false;
25 if (rank[rootX] > rank[rootY]) {
26 parent[rootY] = rootX;
27 } else if (rank[rootX] < rank[rootY]) {
28 parent[rootX] = rootY;
29 } else {
30 parent[rootY] = rootX;
31 rank[rootX]++;
32 }
33 return true;
34 }
35}
36
37public class Solution {
38 public int[] findRedundantDirectedConnection(int[][] edges) {
39 int n = edges.length;
40 int[] parent = new int[n + 1];
41 for (int i = 1; i <= n; i++) {
42 parent[i] = i;
43 }
44
45 int[] can1 = null;
46 int[] can2 = null;
47 for (int[] edge : edges) {
48 int u = edge[0], v = edge[1];
49 if (parent[v] != v) {
50 can1 = new int[] {parent[v], v};
51 can2 = new int[] {u, v};
52 edge[1] = 0;
53 } else {
54 parent[v] = u;
55 }
56 }
57
58 UnionFind uf = new UnionFind(n + 1);
59 for (int[] edge : edges) {
60 if (edge[1] == 0) continue;
61 int u = edge[0], v = edge[1];
62 if (!uf.union(u, v)) return can1 == null ? edge : can1;
63 }
64 return can2;
65 }
66}
67The Java implementation maps nodes with potential redundant edges and explores which connection among candidates could feasibly be removed to preserve tree properties while applying Union-Find for cycle detection, ensuring correct upstream connections converge towards a union of sets strategy.
In this method, we focus on identifying two scenarios: an edge creating a cycle in the graph and a node with two parents. With graph traversal, primarily cross-check with parent pointers and DFS for cycle confirmation, fine-tuning insights to pinpoint a last array occurrence redundant connection.
Time Complexity: O(n^2) with recursive verification.
Space Complexity: O(n), memoizing visited nodes.
1using namespace std;
void dfs(vector<int>& parent, vector<bool>& visited, bool& hasCycle, int node) {
if (visited[node]) {
hasCycle = true;
return;
}
visited[node] = true;
int next = parent[node];
if (next != 0) dfs(parent, visited, hasCycle, next);
visited[node] = false;
}
vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
int n = edges.size();
vector<int> parent(n + 1);
vector<int> can1, can2;
for (auto& edge : edges) {
int u = edge[0], v = edge[1];
if (parent[v] != 0) {
can1 = {parent[v], v};
can2 = edge;
edge[1] = 0;
} else {
parent[v] = u;
}
}
bool hasCycle = false;
for (int i = 1; i <= n; i++) {
vector<bool> visited(n + 1);
dfs(parent, visited, hasCycle, i);
if (hasCycle) break;
}
if (hasCycle) return can1;
return can2;
}
With C++, a recursive DFS uncovers cycles against direct path propositions, accounting for reversed and missed parent elements across the nodes. The final decision uses node connections to wrap redundancies around identified loops effortlessly.