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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
2
Time Complexity: O(N)
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