Find if an array of strings can be chained to form a circle | Set 1

Given an array of strings, find if the given strings can be chained to form a circle. A string X can be put before another string Y in circle if the last character of X is same as first character of Y.

Examples:

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

The idea is to create a directed graph of all characters and then find if their is an eulerian circuit in the graph or not.

Graph representation of some string arrays are given in below diagram,

If there is an eulerian circuit, then chain can be formed, otherwise not.
Note that a directed graph has eulerian circuit only if in degree and out degree of every vertex is same, and all non-zero degree vertices form a single strongly connected component.

Following are detailed steps of the algorithm.

1) Create a directed graph g with number of vertices equal to the size of alphabet. We have created a graph with 26 vertices in the below program.

2) Do following for every string in the given array of strings.
…..a) Add an edge from first character to last character of the given graph.

3) If the created graph has eulerian circuit, then return true, else return false.

Following are C++ and Python implementations of the above algorithm.

C/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
// http://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.
    // Staring 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 'aplha' 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;
}

Python

# 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 xrange(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 xrange(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 xrange(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 xrange(self.V):
            visited[i] = False

        # Do DFS for reversed graph starting from first vertex.
        # Staring 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 xrange(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 xrange(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 xrange(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 xrange(self.V):
            # Recur for all the vertices adjacent to this vertex
            for i in xrange(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 'aplha' edges
    g = Graph(CHARS)

    # Create an edge from first character to last character
    # of every string
    for i in xrange(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


Output:
Can be chained
Can't be chained 

Find if an array of strings can be chained to form a circle | Set 2

This article is contributed by Piyush Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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