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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 constructor(size) {
3 this.parent = Array.from({ length: size }, (_, index) => index);
4 this.rank = Array(size).fill(1);
5 }
6
7 find(x) {
8 if (this.parent[x] !== x) {
9 this.parent[x] = this.find(this.parent[x]);
10 }
11 return this.parent[x];
12 }
13
14 union(x, y) {
15 let rootX = this.find(x);
16 let rootY = this.find(y);
17 if (rootX === rootY) return false;
18 if (this.rank[rootX] > this.rank[rootY]) {
19 this.parent[rootY] = rootX;
20 } else if (this.rank[rootX] < this.rank[rootY]) {
21 this.parent[rootX] = rootY;
22 } else {
23 this.parent[rootY] = rootX;
24 this.rank[rootX]++;
25 }
26 return true;
27 }
28}
29
30var findRedundantDirectedConnection = function(edges) {
31 const n = edges.length;
32 const parent = Array.from({ length: n + 1 }, (_, index) => index);
33 let can1 = null, can2 = null;
34
35 for (const [u, v] of edges) {
36 if (parent[v] !== v) {
37 can1 = [parent[v], v];
38 can2 = [u, v];
39 v = 0;
40 } else {
41 parent[v] = u;
42 }
43 }
44
45 const uf = new UnionFind(n + 1);
46 for (const [u, v] of edges) {
47 if (v === 0) continue;
48 if (!uf.union(u, v)) return can1 || [u, v];
49 }
50 return can2;
51};
52The JavaScript approach involves creating a Union-Find structure tailored to capture duplicate parent incidents and explore the resultant edge arrangements to ensure consistent tree hierarchy by advanced cycle detection and union handling for non-unique parentage correction.
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.