[Leetcode]Graph & Union Find

Graph & Union Find

This time, I will talk about how to use union find to evaluate if there is a cycle in the graph. Union find is a useful method to evaluate the existence of a cycle in an undirected graph, so I will firstly solve a problem by doing so. But actually, this algorithm can even be used to evaluate the existence of a cycle in a directed graph, but it has to satisfied some prerequisites.

[684] Redundant Connection

Description

Analysis

It is easy to understand the problem. What we need to do is to find the last edge leading to the existence of cycle. We can define an array unionSet to record the parents of the node. At first, we initialize the unionSet as unionSets[i] = i. If there is an undirected edge 1-2, then unionSet[unionSet[2]] = 1. The following graph represents a special case. In this case, when edge A comes, unionSet[2] = 1; and then when edge B comes, unionSet[unionSet[2]] = 3. That is, 1, 2 and 3 are connected together. So, if there is an edge, 1-3, we know that unionSet[1] = 3 and unionSets[3] = 3, that means, this two node has already connected together so this edge will lead to the existence of cycle.

Code

#include <iostream>#include <vector>using namespace std;class Solution {private:    int unionFind(int index, vector<int>& sets) {      if (sets[index] == index) {        return index;      } else {        sets[index] = unionFind(sets[index], sets);        return sets[index];      }    }public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {      int size = edges.size();      vector<int> sets(size + 1, 0);      for (int i = 0; i <= size; i++) {        sets[i] = i;      }      vector<int> result;      for (auto each : edges) {        if (unionFind(each[0], sets) == unionFind(each[1], sets)) {          result = each;        } else {          sets[unionFind(each[1], sets)] = unionFind(each[0], sets);        }      }      return result;    }};int main() {  std::vector<std::vector<int>> v{{1,4},{3,4},{1,3},{1,2},{4,5}};  Solution so;  std::vector<int> result = so.findRedundantConnection(v);  cout << result[0] << " " << result[1] << endl;  return 0;}

Time complexity: O(n)

[685] Redundant Connection II

Description

Analysis

At the last problem, the graph is undirected so it will be easy to evaluate the existence of cycle by union find. But now, the graph is directed. The things are still easy.
At first, as the problem’s description said, a rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents, so every node has at most one parent. If a node has two parents, one of its edge connected to its parents should be removed.
Secondly, there shouldn’t be any cycle in the graph.

All in all, if there isn’t any node which has more than 1 parents, then we can just remove the last edge which is leading to the existence of cycle. In this case, we can still use union find because all of the nodes have at most 1 in-degree. If no cycle exists, then we can just remove the second edge which has the same child. What if there are a cycle and the node which has two parents? We have to use a trick. We can set the candidate B invalid, if the cycle doesn’t exist anymore, we know that the candidate B lead to the existence of the cycle. But if the cycle still exists, we know that candidate A lead to the existence of the cycle.

1) Check whether there is a node having two parents.     If so, store them as candidates A and B, and set the second edge invalid. 2) Perform normal union find.     If the tree is now valid            simply return candidate B    else if candidates not existing            we find a circle, return current edge;     else            remove candidate A instead of B.

Code

class Solution {public:    vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {        int n = edges.size();        vector<int> parent(n+1, 0), candA, candB;        // step 1, check whether there is a node with two parents        for (auto &edge:edges) {            if (parent[edge[1]] == 0)                parent[edge[1]] = edge[0];            else {                candA = {parent[edge[1]], edge[1]};                candB = edge;                edge[1] = 0;            }        }        // step 2, union find        for (int i = 1; i <= n; i++) parent[i] = i;        for (auto &edge:edges) {            if (edge[1] == 0) continue;            int u = edge[0], v = edge[1], pu = root(parent, u);            // Now every node only has 1 parent, so root of v is implicitly v            if (pu == v) {                if (candA.empty()) return edge;                return candA;            }            parent[v] = pu;        }        return candB;    }private:    int root(vector<int>& parent, int k) {        if (parent[k] != k)            parent[k] = root(parent, parent[k]);        return parent[k];    }};

Time Complexity: O(n)