# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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> 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> 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> adj = ``new` `Vector>() ; ` `     `  `    ``// create the graph ` `    ``Vector v = ``new` `Vector(); ` `    ``v.add(``0``); ` `    ``v.add(``1``); ` `    ``Vector v1 = ``new` `Vector(); ` `    ``v1.add(``1``); ` `    ``v1.add(``2``); ` `     `  `    ``adj.add(v); ` `    ``adj.add(v1); ` `    ``adj.add(``new` `Vector()); ` `     `  `    ``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> 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> 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> adj = ``new` `List>() ;  ` `     `  `    ``// 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 `

Output:

```2
```

Time Complexity: O(N)

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