Given an undirected graph of N nodes labelled from 1 to N, the task is to find the minimum labelled node that should be removed from the graph such that the resulting graph has no cycle.
Note: If the initial graph has no cycle, i.e. no node needs to be removed, print -1.
Examples:
Input: N = 5, edges[][] = {{5, 1}, {5, 2}, {1, 2}, {2, 3}, {2, 4}}
Output: 1
Explanation:
If node 1 is removed, the resultant graph has no cycle. Similarly, the cycle can be avoided by removing node 2 also.
Since we have to find the minimum labelled node, the answer is 1.Input: N = 5, edges[][] = {{4, 5}, {4, 1}, {4, 2}, {4, 3}, {5, 1}, {5, 2}}
Output: 4
Naive Approach: The naive approach for this problem would be to remove each vertex individually and check whether the resulting graph has a cycle or not. The time complexity for this approach is quadratic.
Efficient Approach: The idea is to apply depth-first search on the given graph and observing the dfs tree formed.
- A Back Edge is referred to as the edge that is not part of the constructed DFS tree and is an edge between some node v and one of the ancestors of v.
- Clearly all those edges of the graph which are not a part of the DFS tree are back edges.
- If there are no back edges in the graph, then the graph has no cycle. So, the answer will be -1 in this case.
If there are back edges in the graph, then we need to find the minimum edge. In order to do this, we need to check if the cycle is removed on removing a specific edge from the graph. Therefore, let v be a vertex which we are currently checking. Therefore, the following conditions must be followed by vertex v such that on removing, it would lead to no cycle:
- v must lie on the tree path connecting the endpoints of each back edge in the graph.
Proof: Suppose there exists some back edge x-y, such that v doesn’t lie on the tree path. If we remove v, we would still be able to traverse from x to y, and back to x via the back edge, indicating that the cycle is not removed. - The subtree of v must have at-most one back edge to any ancestor of v.
Proof: Let the subtree S has to back edges w-x and y-z such that w and y are in S and x and z are ancestors of v. If we remove v, clearly a cycle still exists consisting of the path between w to y, the path from x to z and the two back edges w-x and y-z, i.e. cycle is not removed.
Therefore, the idea is to keep a track of back edges, and an indicator for the number of back edges in the subtree of a node to any of its ancestors. To keep a track of back edges we will use a modified DFS graph colouring algorithm.
In order to check if the subtree v has at-most one back edge to any ancestor of v or not, we implement dfs such that it returns the depth of two highest edges from the subtree of v. We maintain an array where every index ‘i’ in the array stores if the condition 2 from the above is satisfied by the node ‘i’ or not. Similarly, two arrays are implemented, one for the child and another for the parent to see if the node v lies on the tree path connecting the endpoints.
Below is the implementation of the above approach:
C++
// C++ implementation to find the// minimum labelled node to be// removed such that there is no// cycle in the undirected graph#include <bits/stdc++.h>using namespace std;const int MAX = 100005;int totBackEdges;int countAdj[MAX], small[MAX];// Variables to store if a node V has// at-most one back edge and store the// depth of the node for the edgeint isPossible[MAX], depth[MAX];vector<int> adj[MAX];int vis[MAX];// Function to swap the pairs of the graphvoid change(pair<int, int>& p, int x){ // If the second value is // greater than x if (p.second > x) p.second = x; // Put the pair in the ascending // order internally if (p.first > p.second) swap(p.first, p.second);}// Function to perform the DFSpair<int, int> dfs(int v, int p = -1, int de = 0){ // Initialise with the large value pair<int, int> answer(100000000, 100000000); // Storing the depth of this vertex depth[v] = de; // Mark the vertex as visited vis[v] = 1; isPossible[v] = 1; // Iterating through the graph for (int u : adj[v]) { // If the node is a child node if (u ^ p) { // If the child node is unvisited if (!vis[u]) { // Move to the child and increase // the depth auto x = dfs(u, v, de + 1); // increase according to algorithm small[v] += small[u]; change(answer, x.second); change(answer, x.first); // If the node is not having // exactly K backedges if (x.second < de) isPossible[v] = 0; } // If the child is already visited // and in current dfs // (because colour is 1) // then this is a back edge else if (vis[u] == 1) { totBackEdges++; // Increase the countAdj values countAdj[v]++; countAdj[u]++; small[p]++; small[u]--; change(answer, depth[u]); } } } // Colour this vertex 2 as // we are exiting out of // dfs for this node vis[v] = 2; return answer;}// Function to find the minimum labelled// node to be removed such that// there is no cycle in the undirected graphint minNodetoRemove( int n, vector<pair<int, int> > edges){ // Construct the graph for (int i = 0; i < edges.size(); i++) { adj[edges[i].first] .push_back(edges[i].second); adj[edges[i].second] .push_back(edges[i].first); } // Mark visited as false for each node memset(vis, 0, sizeof(vis)); totBackEdges = 0; // Apply dfs on all unmarked nodes for (int v = 1; v <= n; v++) { if (!vis[v]) dfs(v); } // If no backedges in the initial graph // this means that there is no cycle // So, return -1 if (totBackEdges == 0) return -1; int node = -1; // Iterate through the vertices and // return the first node that // satisfies the condition for (int v = 1; v <= n; v++) { // Check whether the count sum of // small[v] and count is the same as // the total back edges and // if the vertex v can be removed if (countAdj[v] + small[v] == totBackEdges && isPossible[v]) { node = v; } if (node != -1) break; } return node;}// Driver codeint main(){ int N = 5; vector<pair<int, int> > edges; edges.push_back(make_pair(5, 1)); edges.push_back(make_pair(5, 2)); edges.push_back(make_pair(1, 2)); edges.push_back(make_pair(2, 3)); edges.push_back(make_pair(2, 4)); cout << minNodetoRemove(N, edges);} |
Java
// Java implementation to find the// minimum labelled node to be// removed such that there is no// cycle in the undirected graphimport java.util.ArrayList;import java.util.Arrays;class Pair { int first, second; public Pair(int first, int second) { this.first = first; this.second = second; }}class GFG{static final int MAX = 100005;static int totBackEdges;static int[] countAdj = new int[MAX];static int[] small = new int[MAX];// Variables to store if a node V has// at-most one back edge and store the// depth of the node for the edgestatic int[] isPossible = new int[MAX];static int[] depth = new int[MAX];@SuppressWarnings("unchecked")static ArrayList<Integer>[] adj = new ArrayList[MAX];static int[] vis = new int[MAX];// Function to swap the pairs of the graphstatic void change(Pair p, int x){ // If the second value is // greater than x if (p.second > x) p.second = x; // Put the Pair in the ascending // order internally if (p.first > p.second) { int tmp = p.first; p.first = p.second; p.second = tmp; }}// Function to perform the DFSstatic Pair dfs(int v, int p, int de){ // Initialise with the large value Pair answer = new Pair(100000000, 100000000); // Storing the depth of this vertex depth[v] = de; // Mark the vertex as visited vis[v] = 1; isPossible[v] = 1; // Iterating through the graph for(int u : adj[v]) { // If the node is a child node if ((u ^ p) != 0) { // If the child node is unvisited if (vis[u] == 0) { // Move to the child and increase // the depth Pair x = dfs(u, v, de + 1); // increase according to algorithm small[v] += small[u]; change(answer, x.second); change(answer, x.first); // If the node is not having // exactly K backedges if (x.second < de) isPossible[v] = 0; } // If the child is already visited // and in current dfs // (because colour is 1) // then this is a back edge else if (vis[u] == 1) { totBackEdges++; // Increase the countAdj values countAdj[v]++; countAdj[u]++; small[p]++; small[u]--; change(answer, depth[u]); } } } // Colour this vertex 2 as // we are exiting out of // dfs for this node vis[v] = 2; return answer;}// Function to find the minimum labelled// node to be removed such that// there is no cycle in the undirected graphstatic int minNodetoRemove(int n, ArrayList<Pair> edges){ // Construct the graph for(int i = 0; i < edges.size(); i++) { adj[edges.get(i).first].add( edges.get(i).second); adj[edges.get(i).second].add( edges.get(i).first); } // Mark visited as false for each node Arrays.fill(vis, 0); totBackEdges = 0; // Apply dfs on all unmarked nodes for(int v = 1; v <= n; v++) { if (vis[v] == 0) dfs(v, -1, 0); } // If no backedges in the initial graph // this means that there is no cycle // So, return -1 if (totBackEdges == 0) return -1; int node = -1; // Iterate through the vertices and // return the first node that // satisfies the condition for(int v = 1; v <= n; v++) { // Check whether the count sum of // small[v] and count is the same as // the total back edges and // if the vertex v can be removed if ((countAdj[v] + small[v] == totBackEdges) && isPossible[v] != 0) { node = v; } if (node != -1) break; } return node;}// Driver codepublic static void main(String[] args) { int N = 5; ArrayList<Pair> edges = new ArrayList<>(); for(int i = 0; i < MAX; i++) { adj[i] = new ArrayList<>(); } edges.add(new Pair(5, 1)); edges.add(new Pair(5, 2)); edges.add(new Pair(1, 2)); edges.add(new Pair(2, 3)); edges.add(new Pair(2, 4)); System.out.println(minNodetoRemove(N, edges));}}// This code is contributed by sanjeev2552 |
Python3
# Python3 implementation to find the# minimum labelled node to be# removed such that there is no# cycle in the undirected graph MAX = 100005; totBackEdges = 0countAdj = [0 for i in range(MAX)]small = [0 for i in range(MAX)] # Variables to store if a node V has# at-most one back edge and store the# depth of the node for the edgeisPossible = [0 for i in range(MAX)]depth = [0 for i in range(MAX)]adj = [[] for i in range(MAX)]vis = [0 for i in range(MAX)]# Function to swap the pairs of the graphdef change(p, x): # If the second value is # greater than x if (p[1] > x): p[1] = x; # Put the pair in the ascending # order internally if (p[0] > p[1]): tmp = p[0]; p[0] = p[1]; p[1] = tmp; # Function to perform the DFSdef dfs(v, p = -1, de = 0): global vis, totBackEdges # Initialise with the large value answer = [100000000, 100000000] # Storing the depth of this vertex depth[v] = de; # Mark the vertex as visited vis[v] = 1; isPossible[v] = 1; # Iterating through the graph for u in adj[v]: # If the node is a child node if ((u ^ p) != 0): # If the child node is unvisited if (vis[u] == 0): # Move to the child and increase # the depth x = dfs(u, v, de + 1); # increase according to algorithm small[v] += small[u]; change(answer, x[1]); change(answer, x[0]); # If the node is not having # exactly K backedges if (x[1] < de): isPossible[v] = 0; # If the child is already visited # and in current dfs # (because colour is 1) # then this is a back edge elif (vis[u] == 1): totBackEdges += 1 # Increase the countAdj values countAdj[v] += 1 countAdj[u] += 1 small[p] += 1 small[u] -= 1 change(answer, depth[u]); # Colour this vertex 2 as # we are exiting out of # dfs for this node vis[v] = 2; return answer; # Function to find the minimum labelled# node to be removed such that# there is no cycle in the undirected graphdef minNodetoRemove( n, edges): # Construct the graph for i in range(len(edges)): adj[edges[i][0]].append(edges[i][1]); adj[edges[i][1]].append(edges[i][0]); global vis, totBackEdges # Mark visited as false for each node vis = [0 for i in range(len(vis))] totBackEdges = 0; # Apply dfs on all unmarked nodes for v in range(1, n + 1): if (vis[v] == 0): dfs(v); # If no backedges in the initial graph # this means that there is no cycle # So, return -1 if (totBackEdges == 0): return -1; node = -1; # Iterate through the vertices and # return the first node that # satisfies the condition for v in range(1, n + 1): # Check whether the count sum of # small[v] and count is the same as # the total back edges and # if the vertex v can be removed if ((countAdj[v] + small[v] == totBackEdges) and isPossible[v] != 0): node = v; if (node != -1): break; return node; # Driver codeif __name__=='__main__': N = 5; edges = [] edges.append([5, 1]); edges.append([5, 2]); edges.append([1, 2]); edges.append([2, 3]); edges.append([2, 4]); print(minNodetoRemove(N, edges)); # This code is contributed by Pratham76 |
C#
// C# implementation to find the// minimum labelled node to be// removed such that there is no// cycle in the undirected graph using System;using System.Collections;using System.Collections.Generic;class GFG{ static int MAX = 100005; static int totBackEdges;static int []countAdj = new int[MAX];static int []small = new int[MAX]; // Variables to store if a node V has// at-most one back edge and store the// depth of the node for the edgestatic int []isPossible = new int[MAX];static int []depth = new int[MAX];static ArrayList adj = new ArrayList(); static int []vis = new int[MAX];class pair{ public int first, second; public pair(int first, int second) { this.first = first; this.second = second; }} // Function to swap the pairs of the graphstatic void change(ref pair p, int x){ // If the second value is // greater than x if (p.second > x) p.second = x; // Put the pair in the ascending // order internally if (p.first > p.second) { int tmp = p.first; p.first = p.second; p.second = tmp; }} // Function to perform the DFSstatic pair dfs(int v, int p = -1, int de = 0){ // Initialise with the large value pair answer = new pair(100000000, 100000000); // Storing the depth of this vertex depth[v] = de; // Mark the vertex as visited vis[v] = 1; isPossible[v] = 1; // Iterating through the graph foreach (int u in (ArrayList)adj[v]) { // If the node is a child node if ((u ^ p) != 0) { // If the child node is unvisited if (vis[u] == 0) { // Move to the child and increase // the depth pair x = dfs(u, v, de + 1); // increase according to algorithm small[v] += small[u]; change(ref answer, x.second); change(ref answer, x.first); // If the node is not having // exactly K backedges if (x.second < de) isPossible[v] = 0; } // If the child is already visited // and in current dfs // (because colour is 1) // then this is a back edge else if (vis[u] == 1) { totBackEdges++; // Increase the countAdj values countAdj[v]++; countAdj[u]++; small[p]++; small[u]--; change(ref answer, depth[u]); } } } // Colour this vertex 2 as // we are exiting out of // dfs for this node vis[v] = 2; return answer;} // Function to find the minimum labelled// node to be removed such that// there is no cycle in the undirected graphstatic int minNodetoRemove( int n, ArrayList edges){ // Construct the graph for (int i = 0; i < edges.Count; i++) { ((ArrayList)adj[((pair)edges[i]).first]) .Add(((pair)edges[i]).second); ((ArrayList)adj[((pair)edges[i]).second]) .Add(((pair)edges[i]).first); } // Mark visited as false for each node Array.Fill(vis, 0); totBackEdges = 0; // Apply dfs on all unmarked nodes for (int v = 1; v <= n; v++) { if (vis[v] == 0) dfs(v); } // If no backedges in the initial graph // this means that there is no cycle // So, return -1 if (totBackEdges == 0) return -1; int node = -1; // Iterate through the vertices and // return the first node that // satisfies the condition for (int v = 1; v <= n; v++) { // Check whether the count sum of // small[v] and count is the same as // the total back edges and // if the vertex v can be removed if ((countAdj[v] + small[v] == totBackEdges) && isPossible[v] != 0) { node = v; } if (node != -1) break; } return node;} // Driver codestatic void Main(){ int N = 5; ArrayList edges = new ArrayList(); for(int i = 0; i < MAX; i++) { adj.Add(new ArrayList()); } edges.Add(new pair(5, 1)); edges.Add(new pair(5, 2)); edges.Add(new pair(1, 2)); edges.Add(new pair(2, 3)); edges.Add(new pair(2, 4)); Console.Write(minNodetoRemove(N, edges)); }}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript implementation to find the// minimum labelled node to be// removed such that there is no// cycle in the undirected graphvar MAX = 100005;var totBackEdges;var countAdj = Array(MAX).fill(0);var small = Array(MAX).fill(0);// Variables to store if a node V has// at-most one back edge and store the// depth of the node for the edgevar isPossible = Array(MAX).fill(0);var depth = Array(MAX).fill(0);var adj = [];var vis = Array(MAX).fill(0);class pair{ constructor(first, second) { this.first = first; this.second = second; }} // Function to swap the pairs of the graphfunction change(p, x){ // If the second value is // greater than x if (p.second > x) p.second = x; // Put the pair in the ascending // order internally if (p.first > p.second) { var tmp = p.first; p.first = p.second; p.second = tmp; }} // Function to perform the DFSfunction dfs(v, p = -1, de = 0){ // Initialise with the large value var answer = new pair(100000000, 100000000); // Storing the depth of this vertex depth[v] = de; // Mark the vertex as visited vis[v] = 1; isPossible[v] = 1; // Iterating through the graph for(var u of adj[v]) { // If the node is a child node if ((u ^ p) != 0) { // If the child node is unvisited if (vis[u] == 0) { // Move to the child and increase // the depth var x = dfs(u, v, de + 1); // Increase according to algorithm small[v] += small[u]; change(answer, x.second); change(answer, x.first); // If the node is not having // exactly K backedges if (x.second < de) isPossible[v] = 0; } // If the child is already visited // and in current dfs // (because colour is 1) // then this is a back edge else if (vis[u] == 1) { totBackEdges++; // Increase the countAdj values countAdj[v]++; countAdj[u]++; small[p]++; small[u]--; change(answer, depth[u]); } } } // Colour this vertex 2 as // we are exiting out of // dfs for this node vis[v] = 2; return answer;} // Function to find the minimum labelled// node to be removed such that// there is no cycle in the undirected graphfunction minNodetoRemove(n, edges){ // Construct the graph for(var i = 0; i < edges.length; i++) { (adj[(edges[i]).first]).push( (edges[i]).second); (adj[(edges[i]).second]).push( (edges[i]).first); } // Mark visited as false for each node vis = Array(MAX).fill(0); totBackEdges = 0; // Apply dfs on all unmarked nodes for(var v = 1; v <= n; v++) { if (vis[v] == 0) dfs(v); } // If no backedges in the initial graph // this means that there is no cycle // So, return -1 if (totBackEdges == 0) return -1; var node = -1; // Iterate through the vertices and // return the first node that // satisfies the condition for(var v = 1; v <= n; v++) { // Check whether the count sum of // small[v] and count is the same as // the total back edges and // if the vertex v can be removed if ((countAdj[v] + small[v] == totBackEdges) && isPossible[v] != 0) { node = v; } if (node != -1) break; } return node;} // Driver codevar N = 5;var edges = [];for(var i = 0; i < MAX; i++){ adj.push(new Array());}edges.push(new pair(5, 1));edges.push(new pair(5, 2));edges.push(new pair(1, 2));edges.push(new pair(2, 3));edges.push(new pair(2, 4));document.write(minNodetoRemove(N, edges));// This code is contributed by rrrtnx</script> |
Output:
1
Time Complexity: O(N + M), where N is the number of nodes and M is the number of edges.
Auxiliary Space: O(N + M).
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