Given a directed tree consisting of N nodes valued from [0, N – 1] and M edges, the task is to find the minimum number of edges that need to reverse for each node X such that there is a path from node X to each vertex of the given Tree.
Examples:
Input: N = 6, edges[][] = {{0, 1}, {1, 3}, {2, 3}, {4, 0}, {4, 5}}
Output: 2 2 2 3 1 2
Explanation:
The answer for node 0 is 2, which can be calculated as:
From 0 to 0: No edges are required to reverse to reach 0 from 0.
From 0 to 1: Can be reached directly using edge 0 -> 1.
From 0 to 2: The edge 2 -> 3 must be reversed for the path 0 -> 1 -> 3 -> 2 to reach 2 from node 0.
From 0 to 3: Can be reached directly as 0 -> 1 -> 3.
From 0 to 4: The edge 4 -> 0 must be reversed to reach 4 from node 0.
From 0 to 5: The edge 4 -> 0 must be reversed to reach 5 from node 0 as 0 -> 4 -> 5To reach every node from the node 0, edge 2 -> 3 and edge 4 -> 0 is reversed. So, a total of 2 edges is reversed for node 0. Similarly, the ans for all the nodes can be calculated.
Input: N = 5, edges[][] = {{1, 0}, {1, 2}, {3, 2}, {3, 4}}
Output: 2 1 2 1 2
Approach: To solve the above problem, the idea is to store the directed edge in the adjacency list along with the reversed directed edge with the negative sign i.e. for directed edge a -> b store the edge a -> b and b -> -a. Then, for each node X of the tree, the answer can be calculated as the number of negative edges encountered in the simple Depth For Search(DFS) from that node X.Follow the steps below to solve the problem:
- Initialize a 2-dimensional vector, say graph[][], to store the edges of the graph.
- Traverse the array edges[][] and for each pair (a, b), push the directed edge b in graph[a] and reversed directed edge -a in graph[b].
-
Iterate over the range N, and for each node:
- Initialize a variable, say count = 0, to count the required number of edges to reverse.
- Call a recursive function, say reorderPaths(node, graph, count, visited) to perform the DFS on the tree.
- Increase the value of count for each of the negative edges traversed in the DFS.
- Print the value of the count, for each iteration of the node.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to perform the DFS traversal // of the tree with reordered paths void reorderPaths(
int s, vector<vector< int > > graph,
int & count, vector< bool >& visited)
{ visited[s] = true ;
// Traverse the adjacency list of
// the source node
for ( auto i : graph[s]) {
if (!visited[ abs (i)]) {
// Reorder the path
if (i < 0)
count++;
// Recursively Call DFS
reorderPaths( abs (i), graph,
count, visited);
}
}
} // Function to find minimum edges to // reverse to make the tree vertices // reachable for each node void minReorder( int n, vector<vector< int > > edges)
{ // Stores the edges
vector<vector< int > > graph(n);
// Traversing the childs
for ( int i = 0; i < edges.size(); i++) {
int a = edges[i][0];
int b = edges[i][1];
// Storing the direct edge
graph[a].push_back(b);
// Storing the reverse edge
graph[b].push_back(-a);
}
// Finding ans for each node
for ( int i = 0; i < n; i++) {
vector< bool > visited(n, false );
int count = 0;
// Function Call
reorderPaths(i, graph, count, visited);
cout << count << " " ;
}
} // Driver Code int main()
{ int N = 6;
vector<vector< int > > edges
= { { 0, 1 }, { 1, 3 }, { 2, 3 }, { 4, 0 }, { 4, 5 } };
minReorder(N, edges);
return 0;
} |
// Java program for the above approach import java.util.*;
public class GFG{
static int count;
// Function to perform the DFS traversal // of the tree with reordered paths static void reorderPaths(
int s, Vector<Integer>[] graph, boolean [] visited)
{ visited[s] = true ;
// Traverse the adjacency list of
// the source node
for ( int i : graph[s]) {
if (!visited[(Math.abs(i))]) {
// Reorder the path
if (i < 0 )
count++;
// Recursively Call DFS
reorderPaths(Math.abs(i), graph, visited);
}
}
} // Function to find minimum edges to // reverse to make the tree vertices // reachable for each node static void minReorder( int n, int [][] edges)
{ // Stores the edges
Vector<Integer>[] graph = new Vector[n];
for ( int i = 0 ; i < n; i++)
graph[i] = new Vector<>();
// Traversing the childs
for ( int i = 0 ; i < edges.length; i++) {
int a = edges[i][ 0 ];
int b = edges[i][ 1 ];
// Storing the direct edge
graph[a].add(b);
// Storing the reverse edge
graph[b].add(-a);
}
// Finding ans for each node
for ( int i = 0 ; i < n; i++) {
boolean []visited = new boolean [n];
count = 0 ;
// Function Call
reorderPaths(i, graph, visited);
System.out.print(count+ " " );
}
} // Driver Code public static void main(String[] args)
{ int N = 6 ;
int [][] edges
= { { 0 , 1 }, { 1 , 3 }, { 2 , 3 }, { 4 , 0 }, { 4 , 5 } };
minReorder(N, edges);
} } // This code is contributed by 29AjayKumar |
# Python 3 program for the above approach count = 0
visited = []
# Function to perform the DFS traversal # of the tree with reordered paths def reorderPaths(s, graph):
global count
visited[s] = True
# Traverse the adjacency list of
# the source node
for i in graph[s]:
if (visited[ abs (i)] = = False ):
# Reorder the path
if (i < 0 ):
count + = 1
# Recursively Call DFS
reorderPaths( abs (i), graph)
# Function to find minimum edges to # reverse to make the tree vertices # reachable for each node def minReorder(n, edges):
global count
global visited
# Stores the edges
graph = [[] for i in range (n)]
# Traversing the childs
for i in range ( len (edges)):
a = edges[i][ 0 ]
b = edges[i][ 1 ]
# Storing the direct edge
graph[a].append(b)
# Storing the reverse edge
graph[b].append( - a)
# Finding ans for each node
for i in range (n):
visited = [ False for i in range (n)]
count = 0
# Function Call
reorderPaths(i, graph)
print (count,end = " " )
# Driver Code if __name__ = = '__main__' :
N = 6
edges = [[ 0 , 1 ],[ 1 , 3 ],[ 2 , 3 ],[ 4 , 0 ],[ 4 , 5 ]]
minReorder(N, edges)
# This code is contributed by SURENDRA_GANGWAR.
|
// C# code for above approach using System;
using System.Collections.Generic;
public class GFG {
static int count;
// Function to perform the DFS traversal
// of the tree with reordered paths
static void ReorderPaths( int s, List< int >[] graph,
bool [] visited)
{
visited[s] = true ;
// Traverse the adjacency list of
// the source node
foreach ( int i in graph[s])
{
if (!visited[(Math.Abs(i))]) {
// Reorder the path
if (i < 0)
count++;
// Recursively Call DFS
ReorderPaths(Math.Abs(i), graph, visited);
}
}
}
// Function to find minimum edges to
// reverse to make the tree vertices
// reachable for each node
static void MinReorder( int n, int [][] edges)
{
// Stores the edges
List< int >[] graph = new List< int >[ n ];
for ( int i = 0; i < n; i++)
graph[i] = new List< int >();
// Traversing the childs
for ( int i = 0; i < edges.Length; i++) {
int a = edges[i][0];
int b = edges[i][1];
// Storing the direct edge
graph[a].Add(b);
// Storing the reverse edge
graph[b].Add(-a);
}
// Finding ans for each node
for ( int i = 0; i < n; i++) {
bool [] visited = new bool [n];
count = 0;
// Function Call
ReorderPaths(i, graph, visited);
Console.Write(count + " " );
}
}
// Driver Code
public static void Main( string [] args)
{
int N = 6;
int [][] edges
= { new int [] { 0, 1 }, new int [] { 1, 3 },
new int [] { 2, 3 }, new int [] { 4, 0 },
new int [] { 4, 5 } };
MinReorder(N, edges);
}
} // This code is contributed by Vaibhav. |
<script> // Javascript program for the above approach
// Function to perform the DFS traversal
// of the tree with reordered paths
let graph;
let edges;
let count;
let visited;
function reorderPaths(s)
{
visited[s] = true ;
// Traverse the adjacency list of
// the source node
for (let i = 0; i < graph[s].length; i++) {
if (!visited[(Math.abs(graph[s][i]))]) {
// Reorder the path
if (graph[s][i] < 0)
count++;
// Recursively Call DFS
reorderPaths(Math.abs(graph[s][i]));
}
}
}
// Function to find minimum edges to
// reverse to make the tree vertices
// reachable for each node
function minReorder(n)
{
// Stores the edges
graph = [];
for (let i = 0; i < n; i++)
{
graph.push([]);
}
// Traversing the childs
for (let i = 0; i < edges.length; i++) {
let a = edges[i][0];
let b = edges[i][1];
// Storing the direct edge
graph[a].push(b);
// Storing the reverse edge
graph[b].push(-a);
}
// Finding ans for each node
for (let i = 0; i < n; i++) {
visited = new Array(n);
visited.fill( false );
count = 0;
// Function Call
reorderPaths(i);
document.write(count + " " );
}
}
let N = 6;
edges = [ [ 0, 1 ], [ 1, 3 ], [ 2, 3 ], [ 4, 0 ], [ 4, 5 ] ];
minReorder(N);
// This code is contributed by decode2207. </script> |
2 2 2 3 1 2
Time Complexity: O(N2)
Auxiliary Space: O(N)