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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.
using namespace std;
class UnionFind {
public:
UnionFind(int size) : parent(size), rank(size, 1) {
for (int i = 0; i < size; ++i) {
parent[i] = i;
}
}
int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
bool unionSets(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) return false;
if (rank[rootX] > rank[rootY]) {
parent[rootY] = rootX;
} else if (rank[rootX] < rank[rootY]) {
parent[rootX] = rootY;
} else {
parent[rootY] = rootX;
rank[rootX]++;
}
return true;
}
private:
vector<int> parent;
vector<int> rank;
};
vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
int n = edges.size();
vector<int> parent(n + 1);
for (int i = 1; i <= n; ++i) {
parent[i] = i;
}
vector<int> can1, can2;
for (const auto& edge : edges) {
int u = edge[0], v = edge[1];
if (parent[v] != v) {
can1 = {parent[v], v};
can2 = {u, v};
edge[1] = 0;
} else {
parent[v] = u;
}
}
UnionFind uf(n + 1);
for (const auto& edge : edges) {
if (edge[1] == 0) continue;
int u = edge[0], v = edge[1];
if (!uf.unionSets(u, v)) return can1.empty() ? edge : can1;
}
return can2;
}
The solution uses a Union-Find class to manage connected components and determine if nodes are already connected, which would indicate a cycle. By capturing candidates that could represent redundant connections, the solution explores potential removal strategies, resolving correctly due to its direction-following and validation process.
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.
1#include <stdio.h>
2#include <stdlib.h>
3#include <stdbool.h>
4
5void dfs(int* parent, bool* visited, bool* cycle, int current) {
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 = true;
13 }
14 }
15}
16
17int* findRedundantDirectedConnection(int** edges, int edgesSize, int* edgesColSize, int* returnSize) {
18 int n = edgesSize;
19 int* parent = (int*)calloc(n + 1, sizeof(int));
20
21 for (int i = 0; i < n; ++i) {
22 int u = edges[i][0], v = edges[i][1];
23 if (parent[v] != 0) {
24 int* can1 = (int*)malloc(sizeof(int) * 2);
25 int* can2 = (int*)malloc(sizeof(int) * 2);
26 can1[0] = parent[v];
27 can1[1] = v;
28 can2[0] = u;
29 can2[1] = v;
30 parent[v] = 0;
31 bool cycle = false;
32 for (int j = 0; j <= n; j++) {
33 if (parent[j] == 0) continue;
34 bool* visited = (bool*)calloc(n + 1, sizeof(bool));
35 dfs(parent, visited, &cycle, j);
36 free(visited);
37 if (cycle) break;
38 }
39 if (cycle) return can1;
40 else return can2;
41 }
42 parent[v] = u;
43 }
44
45 bool cycle = false;
46 for (int j = 0; j <= n; j++) {
47 if (parent[j] == 0) continue;
48 bool* visited = (bool*)calloc(n + 1, sizeof(bool));
49 dfs(parent, visited, &cycle, j);
50 free(visited);
51 if (cycle) {
52 int u = parent[j], v = j;
53 int* result = (int*)malloc(sizeof(int) * 2);
54 result[0] = u;
55 result[1] = v;
56 return result;
57 }
58 }
59
60 return NULL;
61}
62Utilizing DFS, this C code ensures settings target node connections and traces for potential overlapping cycles, using indications to determine redundancy. Understanding traversal in view of directed graph properties is fundamental to isolating missteps.