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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.
The Python code uses the Union-Find construct to handle multiple paths effectively and determines early if redundant connections are present or removal is optimal, readying to designate that action among potentially overlapping connectivity options.
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.
1import java.util.*;
2
3public class Solution {
4 private void dfs(int[] parent, boolean[] visited, boolean[] cycle, int current) {
5 if (cycle[0]) return;
6 visited[current] = true;
7 int next = parent[current];
8 if (next != 0) {
9 if (!visited[next]) {
10 dfs(parent, visited, cycle, next);
11 } else {
12 cycle[0] = true;
13 }
14 }
15 visited[current] = false;
16 }
17
18 public int[] findRedundantDirectedConnection(int[][] edges) {
19 int n = edges.length;
20 int[] parent = new int[n + 1];
21 Arrays.fill(parent, 0);
22
23 int[] can1 = null, can2 = null;
24 for (int[] edge : edges) {
25 int u = edge[0], v = edge[1];
26 if (parent[v] != 0) {
27 can1 = new int[] {parent[v], v};
28 can2 = edge;
29 edge[1] = 0;
30 } else {
31 parent[v] = u;
32 }
33 }
34
35 boolean[] cycle = {false};
36 for (int i = 1; i <= n; i++) {
37 boolean[] visited = new boolean[n + 1];
38 dfs(parent, visited, cycle, i);
39 if (cycle[0]) break;
40 }
41
42 return cycle[0] ? can1 : can2;
43 }
44}
45Java smoothly navigates through the DFS calling structure, ensuring it captures cyclic constructs effectively. With ordered parent review and composite result tones through graph narrative construction, cycles are masterfully exposed.