Find the maximum value permutation of a graph

Given a graph containing N nodes. For any permutation of nodes P1, P2, P3, …, PN the value of the permutation is defined as the number of indices which have at least 1 node on the left of it that has an edge to it. Find the maximum value across all permutations.

Examples:

Input: N = 3, edges[] = {{1, 2}, {2, 3}}
Output: 2
Consider the permutation 2 1 3
Node 1 has node 2 on the left of it and there is an edge connecting them in the graph.
Node 3 has node 2 on the left of it and there is an edge connecting them in the graph.

Input: N = 4, edges[] = {{1, 3}, {2, 4}}
Output: 2
Consider the permutation 1 2 3 4
Node 3 has node 1 on the left of it and there is an edge connecting them in the graph.
Node 4 has node 2 on the left of it and there is an edge connecting them in the graph.

Approach: Let’s start with any node at the beginning, we can follow it up with any node adjacent to it and repeat the process. This resembles a dfs traversal in which every node except the first node has a node before it with which it shares an edge. So for every connected component, the maximum value we can get for the permutation of the nodes of this component is Size of component – 1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the number of nodes
// in the current connected component
int dfs(int x, vector<int> adj[], int vis[])
{
    // Initialise size to 1
    int sz = 1;
  
    // Mark the node as visited
    vis[x] = 1;
  
    // Start a dfs for every unvisited
    // adjacent node
    for (auto ch : adj[x])
        if (!vis[ch])
            sz += dfs(ch, adj, vis);
  
    // Return the number of nodes in
    // the current connected component
    return sz;
}
  
// Function to return the maximum value
// of the required permutation
int maxValue(int n, vector<int> adj[])
{
    int val = 0;
    int vis[n + 1] = { 0 };
  
    // For each connected component
    // add the corresponding value
    for (int i = 1; i <= n; i++)
        if (!vis[i])
            val += dfs(i, adj, vis) - 1;
    return val;
}
  
// Driver code
int main()
{
    int n = 3;
    vector<int> adj[n + 1] = { { 1, 2 }, { 2, 3 } };
    cout << maxValue(n, adj);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
static int vis[];
  
// Function to return the number of nodes
// in the current connected component
static int dfs(int x, Vector<Vector<Integer>> adj)
{
    // Initialise size to 1
    int sz = 1;
  
    // Mark the node as visited
    vis[x] = 1;
  
    // Start a dfs for every unvisited
    // adjacent node
    for (int i = 0; i < adj.get(x).size(); i++)
        if (vis[adj.get(x).get(i)] == 0)
            sz += dfs(adj.get(x).get(i), adj);
  
    // Return the number of nodes in
    // the current connected component
    return sz;
}
  
// Function to return the maximum value
// of the required permutation
static int maxValue(int n, Vector<Vector<Integer>> adj)
{
    int val = 0;
    vis = new int[n + 1];
      
    for (int i = 0; i < n; i++)
    vis[i] = 0;
  
    // For each connected component
    // add the corresponding value
    for (int i = 0; i < n; i++)
        if (vis[i] == 0)
            val += dfs(i, adj) - 1;
    return val;
}
  
// Driver code
public static void main(String args[])
{
    int n = 3;
    Vector<Vector<Integer>> adj = new Vector<Vector<Integer>>() ;
      
    // create the graph
    Vector<Integer> v = new Vector<Integer>();
    v.add(0);
    v.add(1);
    Vector<Integer> v1 = new Vector<Integer>();
    v1.add(1);
    v1.add(2);
      
    adj.add(v);
    adj.add(v1);
    adj.add(new Vector<Integer>());
      
    System.out.println( maxValue(n, adj));
}
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python3 implementation of the approach 
  
# Function to return the number of nodes 
# in the current connected component 
def dfs(x, adj, vis):
  
    # Initialise size to 1 
    sz = 1
  
    # Mark the node as visited 
    vis[x] = 1
  
    # Start a dfs for every unvisited 
    # adjacent node 
    for ch in adj: 
        if (not vis[ch]):
            sz += dfs(ch, adj, vis) 
  
    # Return the number of nodes in 
    # the current connected component 
    return sz 
  
# Function to return the maximum value 
# of the required permutation 
def maxValue(n, adj):
  
    val = 0
    vis = [0] * (n + 1)
  
    # For each connected component 
    # add the corresponding value 
    for i in range(1, n + 1):
        if (not vis[i]):
            val += dfs(i, adj, vis) - 1
    return val 
  
# Driver Code
if __name__ == '__main__':
    n = 3
    adj = [1, 2 , 2, 3]
    print(maxValue(n, adj))
  
# This code is contributed by
# SHUBHAMSINGH10

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C#

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// C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG 
  
static int []vis ; 
  
// Function to return the number of nodes 
// in the current connected component 
static int dfs(int x, List<List<int>> adj) 
    // Initialise size to 1 
    int sz = 1; 
  
    // Mark the node as visited 
    vis[x] = 1; 
  
    // Start a dfs for every unvisited 
    // adjacent node 
    for (int i = 0; i < adj[x].Count; i++) 
        if (vis[adj[x][i]] == 0) 
            sz += dfs(adj[x][i], adj); 
  
    // Return the number of nodes in 
    // the current connected component 
    return sz; 
  
// Function to return the maximum value 
// of the required permutation 
static int maxValue(int n, List<List<int>> adj) 
    int val = 0; 
    vis = new int[n + 1]; 
      
    for (int i = 0; i < n; i++) 
        vis[i] = 0; 
  
    // For each connected component 
    // add the corresponding value 
    for (int i = 0; i < n; i++) 
        if (vis[i] == 0) 
            val += dfs(i, adj) - 1; 
    return val; 
  
// Driver code 
public static void Main(String []args) 
    int n = 3; 
    List<List<int>> adj = new List<List<int>>() ; 
      
    // create the graph 
    List<int> v = new List<int>(); 
    v.Add(0); 
    v.Add(1); 
    List<int> v1 = new List<int>(); 
    v1.Add(1); 
    v1.Add(2); 
      
    adj.Add(v); 
    adj.Add(v1); 
    adj.Add(new List<int>()); 
      
    Console.WriteLine( maxValue(n, adj)); 
  
// This code is contributed by Rajput-Ji

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Output:

2

Time Complexity: O(N)



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