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 there 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.
- 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.
- 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.
- If the created graph has eulerian circuit, then return true, else return false.
Following are C++ and Python implementations of the above algorithm.
// 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 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 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 |
# 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 |
// C# program to check if a given // directed graph is Eulerian or not using System;
using System.Collections.Generic;
// A class that represents an // undirected graph public class GFG {
static readonly int CHARS = 26;
// No. of vertices
int V;
// A dynamic array of adjacency lists
List<List< int > > adj;
int [] ind;
// Constructor
GFG( int V)
{
this .V = V;
ind = new int [V];
adj = new List<List< int > >(CHARS);
for ( int i = 0; i < CHARS; i++) {
adj.Add( new List< int >());
}
}
// Function to add an edge to graph
void addEdge( int v, int w)
{
adj[v].Add(w);
ind[w]++;
}
// Method to check if this graph
// is Eulerian or not
bool 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[i].Count != ind[i])
return false ;
return true ;
}
// This function returns true if all
// non-zero degree vertices of graph
// are strongly connected. Please refer
bool isSC()
{
// Mark all the vertices as not
// visited (For first DFS)
bool [] visited = new bool [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].Count > 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].Count > 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[i].Count > 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, bool [] visited)
{
// Mark the current node as
// visited and print it
visited[v] = true ;
// Recur for all the vertices
// adjacent to this vertex
foreach ( int i in adj[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
foreach ( int i in adj[v])
{
g.adj[i].Add(v);
g.ind[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 bool 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[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 code
static public void Main()
{
String[] arr1 = { "for" , "geek" , "rig" , "kaf" };
int n1 = arr1.Length;
Console.WriteLine((canBeChained(arr1, n1)
? "Can be chained "
: "Can't be chained " ));
String[] arr2 = { "aab" , "abb" };
int n2 = arr2.Length;
Console.WriteLine((canBeChained(arr2, n2)
? "Can be chained "
: "Can't be chained " ));
}
} // This code is contributed by akashish__ |
// A Javascript program to check if // a given directed graph is Eulerian or not let CHARS = 26;
// A class that represents an undirected graph
class Graph {
constructor(V) {
this .V = V; // No. of vertices
this .adj = Array.from(Array(V), () => new Array()); // A dynamic array of adjacency lists
this . in = new Array(V);
for (let i = 0; i < V; i++) {
this . in [i] = 0;
}
}
// function to add an edge to graph
addEdge(v, w) {
this .adj[v].push(w);
this . in [w]++;
}
// Method to check if this graph is Eulerian or not
/* This function returns true if the directed
graph has an eulerian cycle, otherwise returns
false */
isEulerianCycle() {
// Check if all non-zero degree vertices are connected
if ( this .isSC() == false ) return false ;
// Check if in degree and out degree
// of every vertex is same
for (let i = 0; i < this .V; i++)
if ( this .adj[i].length != this . in [i]) return false ;
return true ;
}
// Method/Function to check if all non-zero degree
// vertices are connected
// This function returns true if all non-zero
// degree vertices of graph are strongly connected.
// Please refer
isSC() {
// Mark all the vertices as not visited (For first DFS)
let visited = new Array( this .V);
for (let i = 0; i < this .V; i++) visited[i] = false ;
// Find the first vertex with non-zero degree
let n;
for (n = 0; n < this .V; n++) if ( this .adj[n].length > 0) break ;
// Do DFS traversal starting from first non zero degree vertex.
this .DFSUtil(n, visited);
// If DFS traversal doesn’t visit all vertices, then return false.
for (let i = 0; i < this .V; i++)
if ( this .adj[i].length > 0 && visited[i] == false ) return false ;
// Create a reversed graph
let gr = this .getTranspose();
// Mark all the vertices as not visited (For second DFS)
for (let i = 0; i < this .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 (let i = 0; i < this .V; i++)
if ( this .adj[i].length > 0 && visited[i] == false ) return false ;
return true ;
}
// A recursive function to do DFS starting from v. Used in isConnected();
DFSUtil(v, visited) {
// Mark the current node as visited and print it
visited[v] = true ;
// Recur for all the vertices adjacent to this vertex
for (let j in this .adj[v]) {
let i = this .adj[v][j];
if (!visited[i]) {
this .DFSUtil(i, visited);
}
}
}
// Function that returns reverse (or transpose) of this graph
// This function is needed in isSC()
getTranspose() {
let g = new Graph( this .V);
for (let v = 0; v < this .V; v++) {
// Recur for all the vertices adjacent to this vertex
for (let j in this .adj[v]) {
let i = this .adj[v][j];
g.adj[i].push(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
function canBeChained(arr, n) {
// Create a graph with 'alpha' edges
let g = new Graph(CHARS);
// Create an edge from first character to last character
// of every string
for (let i = 0; i < n; i++) {
let s = arr[i];
g.addEdge(
s[0].charCodeAt() - "a" .charCodeAt(),
s[s.length - 1].charCodeAt() - "a" .charCodeAt()
);
}
// 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
let arr1 = [ "for" , "geek" , "rig" , "kaf" ];
let n1 = arr1.length;
canBeChained(arr1, n1)
? console.log( "Can be chained" )
: console.log( "Can't be chained" );
let arr2 = [ "aab" , "abb" ];
let n2 = arr2.length;
canBeChained(arr2, n2)
? console.log( "Can be chained" )
: console.log( "Can't be chained" );
// This code is contributed by satwiksuman.
|
Can be chained Can't be chained
Time Complexity: O(n*m) where n is max of sizes of arr1 and arr2 and m is maximum size word in arr1 or arr2.
Auxiliary space: O(max(n,m))
Find if an array of strings can be chained to form a circle | Set 2