Given an undirected tree which has even number of vertices, we need to remove the maximum number of edges from this tree such that each connected component of the resultant forest has an even number of vertices.
Examples:
In above shown tree, we can remove at max 2 edges 0-2 and 0-4 shown in red such that each connected component will have even number of vertices.
As we need connected components that have even number of vertices so when we get one component we can remove the edge that connects it to the remaining tree and we will be left with a tree with even number of vertices which will be the same problem but of smaller size, we have to repeat this algorithm until the remaining tree cannot be decomposed further in the above manner.
We will traverse the tree using DFS which will return the number of vertices in the component of which the current node is the root. If a node gets an even number of vertices from one of its children then the edge from that node to its child will be removed and result will be increased by one and if the returned number is odd then we will add it to the number of vertices that the component will have if the current node is the root of it.
1) Do DFS from any starting node as tree is connected. 2) Initialize count of nodes in subtree rooted under current node as 0. 3) Do following recursively for every subtree of current node. a) If size of current subtree is even, increment result by 1 as we can disconnect the subtree. b) Else add count of nodes in current subtree to current count.
Please see below code for better understanding,
/* Program to get maximum number of edges which can be removed such that each connected component
of this tree will have an even number of vertices */
#include <bits/stdc++.h> using namespace std;
// Utility method to do DFS of the graph and count edge // deletion for even forest int dfs(vector< int > g[], int u, bool visit[], int & res)
{ visit[u] = true ;
int currComponentNode = 0;
// iterate over all neighbor of node u
for ( int i = 0; i < g[u].size(); i++)
{
int v = g[u][i];
if (!visit[v])
{
// Count the number of nodes in a subtree
int subtreeNodeCount = dfs(g, v, visit, res);
// if returned node count is even, disconnect
// the subtree and increase result by one.
if (subtreeNodeCount % 2 == 0)
res++;
// else add subtree nodes in current component
else
currComponentNode += subtreeNodeCount;
}
}
// number of nodes in current component and one for
// current node
return (currComponentNode + 1);
} /* method returns max edge that we can remove, after which each connected component will have even number of
vertices */
int maxEdgeRemovalToMakeForestEven(vector< int > g[], int N)
{ // Create a visited array for DFS and make all nodes
// unvisited in starting
bool visit[N + 1];
for ( int i = 0; i <= N; i++)
visit[i] = false ;
int res = 0; // Passed as reference
// calling the dfs from node-0
dfs(g, 0, visit, res);
return res;
} // Utility function to add an undirected edge (u,v) void addEdge(vector< int > g[], int u, int v)
{ g[u].push_back(v);
g[v].push_back(u);
} // Driver code to test above methods int main()
{ int edges[][2] = {{0, 2}, {0, 1}, {0, 4},
{2, 3}, {4, 5}, {5, 6},
{5, 7}};
int N = sizeof (edges)/ sizeof (edges[0]);
vector< int > g[N + 1];
for ( int i = 0; i < N; i++)
addEdge(g, edges[i][0], edges[i][1]);
cout << maxEdgeRemovalToMakeForestEven(g, N);
return 0;
} |
/* Program to get maximum number of edges which can be removed such that each connected component of this tree will have an even number of vertices */ import java.util.*;
class GFG
{ // graph static Vector<Vector<Integer>> g = new Vector<Vector<Integer>>();
static int res;
// Utility method to do DFS of the graph and count edge // deletion for even forest static int dfs( int u, boolean visit[])
{ visit[u] = true ;
int currComponentNode = 0 ;
// iterate over all neighbor of node u
for ( int i = 0 ; i < g.get(u).size(); i++)
{
int v = g.get(u).get(i);
if (!visit[v])
{
// Count the number of nodes in a subtree
int subtreeNodeCount = dfs(v, visit);
// if returned node count is even, disconnect
// the subtree and increase result by one.
if (subtreeNodeCount % 2 == 0 )
res++;
// else add subtree nodes in current component
else
currComponentNode += subtreeNodeCount;
}
}
// number of nodes in current component and one for
// current node
return (currComponentNode + 1 );
} /* method returns max edge that we can remove, after which each connected component will have even number of
vertices */
static int maxEdgeRemovalToMakeForestEven( int N)
{ // Create a visited array for DFS and make all nodes
// unvisited in starting
boolean visit[]= new boolean [N + 1 ];
for ( int i = 0 ; i <= N; i++)
visit[i] = false ;
res = 0 ; // Passed as reference
// calling the dfs from node-0
dfs( 0 , visit);
return res;
} // Utility function to add an undirected edge (u,v) static void addEdge( int u, int v)
{ g.get(u).add(v);
g.get(v).add(u);
} // Driver code to test above methods public static void main(String args[])
{ int edges[][] = {{ 0 , 2 }, { 0 , 1 }, { 0 , 4 },
{ 2 , 3 }, { 4 , 5 }, { 5 , 6 },
{ 5 , 7 }};
int N = edges.length;
for ( int i = 0 ; i < N+ 1 ; i++)
g.add( new Vector<Integer>());
for ( int i = 0 ; i < N; i++)
addEdge( edges[i][ 0 ], edges[i][ 1 ]);
System.out.println(maxEdgeRemovalToMakeForestEven( N));
} } // This code is contributed by Arnab Kundu |
# Python3 program to get maximum number of # edges which can be removed such that each # connected component of this tree will # have an even number of vertices from typing import List
# Utility method to do DFS of the graph # and count edge deletion for even forest def dfs(u: int , visit: List [ bool ]) - > int :
global res, g
visit[u] = True
currComponentNode = 0
# Iterate over all neighbor of node u
for i in range ( len (g[u])):
v = g[u][i]
if ( not visit[v]):
# Count the number of nodes in a subtree
subtreeNodeCount = dfs(v, visit)
# If returned node count is even, disconnect
# the subtree and increase result by one.
if (subtreeNodeCount % 2 = = 0 ):
res + = 1
# Else add subtree nodes in
# current component
else :
currComponentNode + = subtreeNodeCount
# Number of nodes in current component
# and one for current node
return (currComponentNode + 1 )
# Method returns max edge that we can remove, # after which each connected component will # have even number of vertices def maxEdgeRemovalToMakeForestEven(N: int ) - > int :
# Create a visited array for DFS and make
# all nodes unvisited in starting
visit = [ False for _ in range (N + 1 )]
# Calling the dfs from node-0
dfs( 0 , visit)
return res
# Utility function to add an undirected edge (u,v) def addEdge(u: int , v: int ) - > None :
global g
g[u].append(v)
g[v].append(u)
# Driver code if __name__ = = "__main__" :
res = 0
edges = [ [ 0 , 2 ], [ 0 , 1 ],
[ 0 , 4 ], [ 2 , 3 ],
[ 4 , 5 ], [ 5 , 6 ],
[ 5 , 7 ] ]
N = len (edges)
g = [[] for _ in range (N + 1 )]
for i in range (N):
addEdge(edges[i][ 0 ], edges[i][ 1 ])
print (maxEdgeRemovalToMakeForestEven(N))
# This code is contributed by sanjeev2552 |
/* C# Program to get maximum number of edges which can be removed such that each connected component of this tree will have an even number of vertices */ using System;
using System.Collections.Generic;
class GFG
{ // graph static List<List< int >> g = new List<List< int >>();
static int res;
// Utility method to do DFS of the graph and // count edge deletion for even forest static int dfs( int u, bool []visit)
{ visit[u] = true ;
int currComponentNode = 0;
// iterate over all neighbor of node u
for ( int i = 0; i < g[u].Count; i++)
{
int v = g[u][i];
if (!visit[v])
{
// Count the number of nodes in a subtree
int subtreeNodeCount = dfs(v, visit);
// if returned node count is even, disconnect
// the subtree and increase result by one.
if (subtreeNodeCount % 2 == 0)
res++;
// else add subtree nodes in current component
else
currComponentNode += subtreeNodeCount;
}
}
// number of nodes in current component and one for
// current node
return (currComponentNode + 1);
} /* method returns max edge that we can remove, after which each connected component will have even number of vertices */ static int maxEdgeRemovalToMakeForestEven( int N)
{ // Create a visited array for DFS and
// make all nodes unvisited in starting
bool []visit = new bool [N + 1];
for ( int i = 0; i <= N; i++)
visit[i] = false ;
res = 0; // Passed as reference
// calling the dfs from node-0
dfs(0, visit);
return res;
} // Utility function to add an undirected edge (u,v) static void addEdge( int u, int v)
{ g[u].Add(v);
g[v].Add(u);
} // Driver code public static void Main(String []args)
{ int [,]edges = {{0, 2}, {0, 1}, {0, 4},
{2, 3}, {4, 5}, {5, 6},
{5, 7}};
int N = edges.GetLength(0);
for ( int i = 0; i < N + 1; i++)
g.Add( new List< int >());
for ( int i = 0; i < N; i++)
addEdge(edges[i, 0], edges[i, 1]);
Console.WriteLine(maxEdgeRemovalToMakeForestEven(N));
} } // This code is contributed by 29AjayKumar |
<script> /* Program to get maximum number of edges which can be removed such that each connected component of this tree will have an even number of vertices */ // graph let g = []; let res; // Utility method to do DFS of the graph and count edge // deletion for even forest function dfs(u,visit)
{ visit[u] = true ;
let currComponentNode = 0;
// iterate over all neighbor of node u
for (let i = 0; i < g[u].length; i++)
{
let v = g[u][i];
if (!visit[v])
{
// Count the number of nodes in a subtree
let subtreeNodeCount = dfs(v, visit);
// if returned node count is even, disconnect
// the subtree and increase result by one.
if (subtreeNodeCount % 2 == 0)
res++;
// else add subtree nodes in current component
else
currComponentNode += subtreeNodeCount;
}
}
// number of nodes in current component and one for
// current node
return (currComponentNode + 1);
} /* method returns max edge that we can remove, after which each connected component will have even number of
vertices */
function maxEdgeRemovalToMakeForestEven(N)
{ // Create a visited array for DFS and make all nodes
// unvisited in starting
let visit= new Array(N + 1);
for (let i = 0; i <= N; i++)
visit[i] = false ;
res = 0; // Passed as reference
// calling the dfs from node-0
dfs(0, visit);
return res;
} // Utility function to add an undirected edge (u,v) function addEdge(u,v)
{ g[u].push(v);
g[v].push(u);
} // Driver code to test above methods let edges=[[0, 2], [0, 1], [0, 4], [2, 3], [4, 5], [5, 6],
[5, 7]];
let N = edges.length; for (let i = 0; i < N+1; i++)
g.push([]);
for (let i = 0; i < N; i++)
addEdge( edges[i][0], edges[i][1]);
document.write(maxEdgeRemovalToMakeForestEven( N)); // This code is contributed by rag2127 </script> |
Output:
2
Time Complexity: O(n) where n is number of nodes in tree.
Auxiliary Space: O(n)