Encuentra si una array de strings se puede enstringr para formar un círculo | Serie 1

Dada una serie de strings, encuentre si las strings dadas se pueden enstringr para formar un círculo. Una string X se puede poner antes de otra string Y en un círculo si el último carácter de X es el mismo que el primer carácter de Y.
Ejemplos: 

Input: arr[] = {"geek", "king"}
Output: Yes, the given strings can be chained.
Note that the last character of first string is same
as first character of second string and vice versa is
also true.

Input: arr[] = {"for", "geek", "rig", "kaf"}
Output: Yes, the given strings can be chained.
The strings can be chained as "for", "rig", "geek" 
and "kaf"

Input: arr[] = {"aab", "bac", "aaa", "cda"}
Output: Yes, the given strings can be chained.
The strings can be chained as "aaa", "aab", "bac" 
and "cda"

Input: arr[] = {"aaa", "bbb", "baa", "aab"};
Output: Yes, the given strings can be chained.
The strings can be chained as "aaa", "aab", "bbb" 
and "baa"

Input: arr[] = {"aaa"};
Output: Yes

Input: arr[] = {"aaa", "bbb"};
Output: No

Input  : arr[] = ["abc", "efg", "cde", "ghi", "ija"]
Output : Yes
These strings can be reordered as, “abc”, “cde”, “efg”,
“ghi”, “ija”

Input : arr[] = [“ijk”, “kji”, “abc”, “cba”]
Output : No

La idea es crear un gráfico dirigido de todos los caracteres y luego encontrar si hay un circuito euleriano en el gráfico o no. 
La representación gráfica de algunas arrays de strings se proporciona en el siguiente diagrama, 

Si hay un circuito euleriano , entonces se puede formar una string, de lo contrario no. 

Tenga en cuenta que un gráfico dirigido tiene un circuito euleriano solo si el grado de entrada y salida de cada vértice es el mismo, y todos los vértices de grado distinto de cero forman un solo componente fuertemente conectado.

Los siguientes son pasos detallados del algoritmo.

  1. Crea un gráfico dirigido g con un número de vértices igual al tamaño del alfabeto. Hemos creado un gráfico con 26 vértices en el siguiente programa.
  2. Haga lo siguiente para cada string en la array de strings dada. 
    • …..a) Agregue un borde desde el primer carácter hasta el último carácter del gráfico dado.
  3. Si el gráfico creado tiene un circuito euleriano , devuelve verdadero, de lo contrario, devuelve falso.

Las siguientes son implementaciones en C++ y Python del algoritmo anterior.

C++

// A C++ program to check if a given
// directed graph is Eulerian or not
#include<iostream>
#include <list>
#define CHARS 26
using namespace std;
 
// A class that represents an undirected graph
class Graph
{
    int V;    // No. of vertices
    list<int> *adj; // A dynamic array of adjacency lists
    int *in;
public:
    // Constructor and destructor
    Graph(int V);
    ~Graph()   { delete [] adj; delete [] in; }
 
    // function to add an edge to graph
    void addEdge(int v, int w) { adj[v].push_back(w);  (in[w])++; }
 
    // Method to check if this graph is Eulerian or not
    bool isEulerianCycle();
 
    // Method to check if all non-zero degree
    // vertices are connected
    bool isSC();
 
    // Function to do DFS starting from v. Used in isConnected();
    void DFSUtil(int v, bool visited[]);
 
    Graph getTranspose();
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
    in = new int[V];
    for (int i = 0; i < V; i++)
       in[i] = 0;
}
 
/* This function returns true if the directed
   graph has an eulerian cycle, otherwise returns
   false  */
bool Graph::isEulerianCycle()
{
    // Check if all non-zero degree vertices are connected
    if (isSC() == false)
        return false;
 
    // Check if in degree and out degree
    // of every vertex is same
    for (int i = 0; i < V; i++)
        if (adj[i].size() != in[i])
            return false;
 
    return true;
}
 
// A recursive function to do DFS starting from v
void Graph::DFSUtil(int v, bool visited[])
{
    // Mark the current node as visited and print it
    visited[v] = true;
 
    // Recur for all the vertices adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            DFSUtil(*i, visited);
}
 
// Function that returns reverse (or transpose) of this graph
// This function is needed in isSC()
Graph Graph::getTranspose()
{
    Graph g(V);
    for (int v = 0; v < V; v++)
    {
        // Recur for all the vertices adjacent to this vertex
        list<int>::iterator i;
        for(i = adj[v].begin(); i != adj[v].end(); ++i)
        {
            g.adj[*i].push_back(v);
            (g.in[v])++;
        }
    }
    return g;
}
 
// This function returns true if all non-zero
// degree vertices of graph are strongly connected.
// Please refer
// https://www.geeksforgeeks.org/connectivity-in-a-directed-graph/
bool Graph::isSC()
{
    // Mark all the vertices as not visited (For first DFS)
    bool visited[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
    // Find the first vertex with non-zero degree
    int n;
    for (n = 0; n < V; n++)
        if (adj[n].size() > 0)
          break;
 
    // Do DFS traversal starting from first non zero degree vertex.
    DFSUtil(n, visited);
 
     // If DFS traversal doesn’t visit all vertices, then return false.
    for (int i = 0; i < V; i++)
        if (adj[i].size() > 0 && visited[i] == false)
              return false;
 
    // Create a reversed graph
    Graph gr = getTranspose();
 
    // Mark all the vertices as not visited (For second DFS)
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
    // Do DFS for reversed graph starting from first vertex.
    // Starting Vertex must be same starting point of first DFS
    gr.DFSUtil(n, visited);
 
    // If all vertices are not visited in second DFS, then
    // return false
    for (int i = 0; i < V; i++)
        if (adj[i].size() > 0 && visited[i] == false)
             return false;
 
    return true;
}
 
// This function takes an of strings and returns true
// if the given array of strings can be chained to
// form cycle
bool canBeChained(string arr[], int n)
{
    // Create a graph with 'alpha' edges
    Graph g(CHARS);
 
    // Create an edge from first character to last character
    // of every string
    for (int i = 0; i < n; i++)
    {
        string s = arr[i];
        g.addEdge(s[0]-'a', s[s.length()-1]-'a');
    }
 
    // The given array of strings can be chained if there
    // is an eulerian cycle in the created graph
    return g.isEulerianCycle();
}
 
// Driver program to test above functions
int main()
{
    string arr1[] =  {"for", "geek", "rig", "kaf"};
    int n1 = sizeof(arr1)/sizeof(arr1[0]);
    canBeChained(arr1, n1)?  cout << "Can be chained \n" :
                           cout << "Can't be chained \n";
 
    string arr2[] =  {"aab", "abb"};
    int n2 = sizeof(arr2)/sizeof(arr2[0]);
    canBeChained(arr2, n2)?  cout << "Can be chained \n" :
                           cout << "Can't be chained \n";
 
    return 0;
}

Java

// Java program to check if a given
// directed graph is Eulerian or not
import java.util.ArrayList;
import java.util.List;
 
// A class that represents an
// undirected graph
class GFG{
     
static final int CHARS = 26;
 
// No. of vertices
int V;
 
// A dynamic array of adjacency lists
List<List<Integer>> adj;
int[] in;
 
// Constructor
GFG(int V)
{
    this.V = V;
    in = new int[V];
    adj = new ArrayList<>(CHARS);
     
    for(int i = 0; i < CHARS; i++)
    {
       adj.add(i, new ArrayList<>());
    }
}
 
// Function to add an edge to graph
void addEdge(int v, int w)
{
    adj.get(v).add(w);
    in[w]++;
}
 
// Method to check if this graph
// is Eulerian or not
boolean isEulerianCycle()
{
     
    // Check if all non-zero degree
    // vertices are connected
    if (!isSC())
        return false;
 
    // Check if in degree and out
    // degree of every vertex is same
    for(int i = 0; i < V; i++)
       if (adj.get(i).size() != in[i])
           return false;
 
    return true;
}
 
// This function returns true if all
// non-zero degree vertices of graph
// are strongly connected. Please refer
boolean isSC()
{
     
    // Mark all the vertices as not
    // visited (For first DFS)
    boolean[] visited = new boolean[V];
    for(int i = 0; i < V; i++)
       visited[i] = false;
 
    // Find the first vertex with
    // non-zero degree
    int n;
    for(n = 0; n < V; n++)
       if (adj.get(n).size() > 0)
           break;
 
    // Do DFS traversal starting from
    // first non zero degree vertex.
    DFSUtil(n, visited);
 
    // If DFS traversal doesn't visit all
    // vertices, then return false.
    for(int i = 0; i < V; i++)
       if (adj.get(i).size() > 0 && !visited[i])
           return false;
 
    // Create a reversed graph
    GFG gr = getTranspose();
 
    // Mark all the vertices as not
    // visited (For second DFS)
    for(int i = 0; i < V; i++)
       visited[i] = false;
 
    // Do DFS for reversed graph starting
    // from first vertex. Starting Vertex
    // must be same starting point of first DFS
    gr.DFSUtil(n, visited);
 
    // If all vertices are not visited in
    // second DFS, then return false
    for(int i = 0; i < V; i++)
       if (adj.get(i).size() > 0 && !visited[i])
           return false;
 
    return true;
}
 
// Function to do DFS starting from v.
// Used in isConnected();
// A recursive function to do DFS
// starting from v
void DFSUtil(int v, boolean[] visited)
{
     
    // Mark the current node as
    // visited and print it
    visited[v] = true;
 
    // Recur for all the vertices
    // adjacent to this vertex
    for(Integer i : adj.get(v))
       if (!visited[i])
       {
           DFSUtil(i, visited);
       }
}
 
// Function that returns reverse
// (or transpose) of this graph
// This function is needed in isSC()
GFG getTranspose()
{
    GFG g = new GFG(V);
    for(int v = 0; v < V; v++)
    {
        
       // Recur for all the vertices
       // adjacent to this vertex
       for(Integer i : adj.get(v))
       {
          g.adj.get(i).add(v);
          g.in[v]++;
       }
    }
    return g;
}
 
// This function takes an of strings
// and returns true if the given array
// of strings can be chained to form cycle
static boolean canBeChained(String[] arr, int n)
{
     
    // Create a graph with 'alpha' edges
    GFG g = new GFG(CHARS);
 
    // Create an edge from first character
    // to last character of every string
    for(int i = 0; i < n; i++)
    {
       String s = arr[i];
       g.addEdge(s.charAt(0) - 'a',
                 s.charAt(s.length() - 1) - 'a');
    }
     
    // The given array of strings can be
    // chained if there is an eulerian
    // cycle in the created graph
    return g.isEulerianCycle();
}
 
// Driver code
public static void main(String[] args) throws Exception
{
    String[] arr1 = { "for", "geek",
                      "rig", "kaf" };
    int n1 = arr1.length;
     
    System.out.println((canBeChained(arr1, n1) ?
                       "Can be chained " :
                       "Can't be chained "));
 
    String[] arr2 = { "aab", "abb" };
    int n2 = arr2.length;
     
    System.out.println((canBeChained(arr2, n2) ?
                       "Can be chained " :
                       "Can't be chained "));
}
}
 
// This code is contributed by abhay379201

Python3

# Python program to check if a given directed graph is Eulerian or not
CHARS = 26
 
# A class that represents an undirected graph
class Graph(object):
    def __init__(self, V):
        self.V = V      # No. of vertices
        self.adj = [[] for x in range(V)]  # a dynamic array
        self.inp = [0] * V
 
    # function to add an edge to graph
    def addEdge(self, v, w):
        self.adj[v].append(w)
        self.inp[w]+=1
 
    # Method to check if this graph is Eulerian or not
    def isSC(self):
        # Mark all the vertices as not visited (For first DFS)
        visited = [False] * self.V
 
        # Find the first vertex with non-zero degree
        n = 0
        for n in range(self.V):
            if len(self.adj[n]) > 0:
                break
 
        # Do DFS traversal starting from first non zero degree vertex.
        self.DFSUtil(n, visited)
 
        # If DFS traversal doesn't visit all vertices, then return false.
        for i in range(self.V):
            if len(self.adj[i]) > 0 and visited[i] == False:
                return False
 
        # Create a reversed graph
        gr = self.getTranspose()
 
        # Mark all the vertices as not visited (For second DFS)
        for i in range(self.V):
            visited[i] = False
 
        # Do DFS for reversed graph starting from first vertex.
        # Starting Vertex must be same starting point of first DFS
        gr.DFSUtil(n, visited)
 
        # If all vertices are not visited in second DFS, then
        # return false
        for i in range(self.V):
            if len(self.adj[i]) > 0 and visited[i] == False:
                return False
 
        return True
 
    # This function returns true if the directed graph has an eulerian
    # cycle, otherwise returns false
    def isEulerianCycle(self):
 
        # Check if all non-zero degree vertices are connected
        if self.isSC() == False:
            return False
 
        # Check if in degree and out degree of every vertex is same
        for i in range(self.V):
            if len(self.adj[i]) != self.inp[i]:
                return False
 
        return True
 
    # A recursive function to do DFS starting from v
    def DFSUtil(self, v, visited):
 
        # Mark the current node as visited and print it
        visited[v] = True
 
        # Recur for all the vertices adjacent to this vertex
        for i in range(len(self.adj[v])):
            if not visited[self.adj[v][i]]:
                self.DFSUtil(self.adj[v][i], visited)
 
    # Function that returns reverse (or transpose) of this graph
    # This function is needed in isSC()
    def getTranspose(self):
        g = Graph(self.V)
        for v in range(self.V):
            # Recur for all the vertices adjacent to this vertex
            for i in range(len(self.adj[v])):
                g.adj[self.adj[v][i]].append(v)
                g.inp[v]+=1
        return g
 
# This function takes an of strings and returns true
# if the given array of strings can be chained to
# form cycle
def canBeChained(arr, n):
 
    # Create a graph with 'alpha' edges
    g = Graph(CHARS)
 
    # Create an edge from first character to last character
    # of every string
    for i in range(n):
        s = arr[i]
        g.addEdge(ord(s[0])-ord('a'), ord(s[len(s)-1])-ord('a'))
 
    # The given array of strings can be chained if there
    # is an eulerian cycle in the created graph
    return g.isEulerianCycle()
 
# Driver program
arr1 = ["for", "geek", "rig", "kaf"]
n1 = len(arr1)
if canBeChained(arr1, n1):
    print ("Can be chained")
else:
    print ("Cant be chained")
 
arr2 = ["aab", "abb"]
n2 = len(arr2)
if canBeChained(arr2, n2):
    print ("Can be chained")
else:
    print ("Can't be chained")
 
# This code is contributed by BHAVYA JAIN
Producción

Can be chained 
Can't be chained 

Encuentra si una array de strings se puede enstringr para formar un círculo | conjunto 2 

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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