Arrange array elements such that last digit of an element is equal to first digit of the next element
Given an array arr[] of integers, the task is to arrange the array elements such that the last digit of an element is equal to first digit of the next element.
Examples:
Input: arr[] = {123, 321}
Output: 123 321Input: arr[] = {451, 378, 123, 1254}
Output: 1254 451 123 378
Naive approach: Find all the permutations of the array elements and then print the arranged array which meets the required condition. The time complexity of this approach is O(N!)
Efficient approach: Create a directed graph where there will be a directed edge from a node A to node B if the last digit of the number represented by Node A is equal to the first digit of the number represented by Node B. Now, find the Eulerian path for the graph formed. The complexity of the above algorithm is O(E * E) where E is the number of edges in the graph.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // To store the array elements vector<string> arr; // Adjacency list for the graph nodes vector<vector< int > > graph; // To store the euler path vector<string> path; // Print eulerian path bool print_euler( int i, int visited[], int count) { // Mark node as visited // and increase the count visited[i] = 1; count++; // If all the nodes are visited // then we have traversed the euler path if (count == graph.size()) { path.push_back(arr[i]); return true ; } // Check if the node lies in euler path bool b = false ; // Traverse through remaining edges for ( int j = 0; j < graph[i].size(); j++) if (visited[graph[i][j]] == 0) { b |= print_euler(graph[i][j], visited, count); } // If the euler path is found if (b) { path.push_back(arr[i]); return true ; } // Else unmark the node else { visited[i] = 0; count--; return false ; } } // Function to create the graph and // print the required path void connect() { int n = arr.size(); graph.clear(); graph.resize(n); // Connect the nodes for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { if (i == j) continue ; // If the last character matches with the // first character if (arr[i][arr[i].length() - 1] == arr[j][0]) { graph[i].push_back(j); } } } // Print the path for ( int i = 0; i < n; i++) { int visited[n] = { 0 }, count = 0; // If the euler path starts // from the ith node if (print_euler(i, visited, count)) break ; } // Print the euler path for ( int i = path.size() - 1; i >= 0; i--) { cout << path[i]; if (i != 0) cout << " " ; } } // Driver code int main() { arr.push_back( "451" ); arr.push_back( "378" ); arr.push_back( "123" ); arr.push_back( "1254" ); // Create graph and print the path connect(); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG{ // To store the array elements static List<String> arr = new ArrayList<String>(); // Adjacency list for the graph nodes static List<List<Integer>> graph = new ArrayList<List<Integer>>(); // To store the euler path static List<String> path = new ArrayList<String>(); // Print eulerian path static boolean print_euler( int i, int []visited, int count) { // Mark node as visited // and increase the count visited[i] = 1 ; count++; // If all the nodes are visited // then we have traversed the euler path if (count == graph.size()) { path.add(arr.get(i)); return true ; } // Check if the node lies in euler path boolean b = false ; // Traverse through remaining edges for ( int j = 0 ; j < graph.get(i).size(); j++) if (visited[graph.get(i).get(j)] == 0 ) { b |= print_euler(graph.get(i).get(j), visited, count); } // If the euler path is found if (b) { path.add(arr.get(i)); return true ; } // Else unmark the node else { visited[i] = 0 ; count--; return false ; } } // Function to create the graph and // print the required path static void connect() { int n = arr.size(); graph = new ArrayList<List<Integer>>(n); for ( int i = 0 ; i < n; i++) { graph.add( new ArrayList<Integer>()); } // Connect the nodes for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { if (i == j) continue ; // If the last character matches with the // first character if (arr.get(i).charAt((arr.get(i).length()) - 1 ) == arr.get(j).charAt( 0 )) { graph.get(i).add(j); } } } // Print the path for ( int i = 0 ; i < n; i++) { int []visited = new int [n]; int count = 0 ; // If the euler path starts // from the ith node if (print_euler(i, visited, count)) break ; } // Print the euler path for ( int i = path.size() - 1 ; i >= 0 ; i--) { System.out.print(path.get(i)); if (i != 0 ) System.out.print( " " ); } } // Driver code public static void main(String []args) { arr.add( "451" ); arr.add( "378" ); arr.add( "123" ); arr.add( "1254" ); // Create graph and print the path connect(); } } // This code is contributed by pratham76 |
Python3
# Python3 implementation of the approach # Print eulerian path def print_euler(i, visited, count): # Mark node as visited # and increase the count visited[i] = 1 count + = 1 # If all the nodes are visited then # we have traversed the euler path if count = = len (graph): path.append(arr[i]) return True # Check if the node lies in euler path b = False # Traverse through remaining edges for j in range ( 0 , len (graph[i])): if visited[graph[i][j]] = = 0 : b | = print_euler(graph[i][j], visited, count) # If the euler path is found if b: path.append(arr[i]) return True # Else unmark the node else : visited[i] = 0 count - = 1 return False # Function to create the graph # and print the required path def connect(): n = len (arr) # Connect the nodes for i in range ( 0 , n): for j in range ( 0 , n): if i = = j: continue # If the last character matches # with the first character if arr[i][ - 1 ] = = arr[j][ 0 ]: graph[i].append(j) # Print the path for i in range ( 0 , n): visited = [ 0 ] * n count = 0 # If the euler path starts # from the ith node if print_euler(i, visited, count): break # Print the euler path for i in range ( len (path) - 1 , - 1 , - 1 ): print (path[i], end = "") if i ! = 0 : print ( " " , end = "") # Driver code if __name__ = = "__main__" : # To store the array elements arr = [] arr.append( "451" ) arr.append( "378" ) arr.append( "123" ) arr.append( "1254" ) # Adjacency list for the graph nodes graph = [[] for i in range ( len (arr))] # To store the euler path path = [] # Create graph and print the path connect() # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approach using System; using System.Collections; using System.Collections.Generic; class GFG { // To store the array elements static List< string > arr = new List< string >(); // Adjacency list for the graph nodes static List<List< int > > graph= new List<List< int >>(); // To store the euler path static List< string > path = new List< string >(); // Print eulerian path static bool print_euler( int i, int []visited, int count) { // Mark node as visited // and increase the count visited[i] = 1; count++; // If all the nodes are visited // then we have traversed the euler path if (count == graph.Count) { path.Add(arr[i]); return true ; } // Check if the node lies in euler path bool b = false ; // Traverse through remaining edges for ( int j = 0; j < graph[i].Count; j++) if (visited[graph[i][j]] == 0) { b |= print_euler(graph[i][j], visited, count); } // If the euler path is found if (b) { path.Add(arr[i]); return true ; } // Else unmark the node else { visited[i] = 0; count--; return false ; } } // Function to create the graph and // print the required path static void connect() { int n = arr.Count; graph= new List<List< int >>(n); for ( int i = 0; i < n; i++) { graph.Add( new List< int >()); } // Connect the nodes for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { if (i == j) continue ; // If the last character matches with the // first character if (arr[i][(arr[i].Length) - 1] == arr[j][0]) { graph[i].Add(j); } } } // Print the path for ( int i = 0; i < n; i++) { int []visited = new int [n]; int count = 0; // If the euler path starts // from the ith node if (print_euler(i, visited, count)) break ; } // Print the euler path for ( int i = path.Count - 1; i >= 0; i--) { Console.Write(path[i]); if (i != 0) Console.Write( " " ); } } // Driver code public static void Main( params string []args) { arr.Add( "451" ); arr.Add( "378" ); arr.Add( "123" ); arr.Add( "1254" ); // Create graph and print the path connect(); } } // This code is contributed by rutvik_56. |
Javascript
// JavaScript implementation of the approach // To store the array elements let arr = []; // Adjacency list for the graph nodes let graph = []; // To store the euler path let path = []; // Print eulerian path function print_euler(i, visited, count) { // Mark node as visited // and increase the count visited[i] = 1; count++; // If all the nodes are visited // then we have traversed the euler path if (count == graph.length) { path.push(arr[i]); return true ; } // Check if the node lies in euler path let b = false ; // Traverse through remaining edges for (let j = 0; j < graph[i].length; j++) if (visited[graph[i][j]] == 0) { b |= print_euler(graph[i][j], visited, count); } // If the euler path is found if (b) { path.push(arr[i]); return true ; } // Else unmark the node else { visited[i] = 0; count--; return false ; } } // Function to create the graph and // print the required path function connect() { let n = arr.length; graph.length = 0; graph.length = n; // Connect the nodes for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { if (i == j) continue ; // If the last character matches with the // first character if (arr[i][arr[i].length - 1] == arr[j][0]) { graph[i].push(j); } } } // Print the path for (let i = 0; i < n; i++) { let visited = new Array(n).fill(0); let count = 0; // If the euler path starts // from the ith node if (print_euler(i, visited, count)) break ; } // Print the euler path for (let i = path.length - 1; i >= 0; i--) { console.log(path[i]); if (i != 0) console.log( " " ); } } // Driver code ( function () { arr.push( "451" ); arr.push( "378" ); arr.push( "123" ); arr.push( "1254" ); // Create graph and print the path connect(); })(); |
1254 451 123 378
Time Complexity : O(N* log(N))
Auxiliary Space: O(N)
Approach:(graph-based)
The approach is to create a graph where each number is a node and add an edge between two nodes if the last digit of one number matches the first digit of the other number. Then, we use an algorithm to find the Eulerian path in the graph, which represents the valid arrangement of numbers. Finally, we print the Eulerian path in reverse order to obtain the required arrangement of numbers.
C++
#include <bits/stdc++.h> using namespace std; // To store the array elements vector<string> arr; // Adjacency list for the graph nodes vector<vector< int >> graph; // To store the euler path vector<string> path; // Print eulerian path bool print_euler( int i, int visited[], int count) { // Mark node as visited // and increase the count visited[i] = 1; count++; // If all the nodes are visited // then we have traversed the euler path if (count == graph.size()) { path.push_back(arr[i]); return true ; } // Check if the node lies in euler path bool b = false ; // Traverse through remaining edges for ( int j = 0; j < graph[i].size(); j++) { if (visited[graph[i][j]] == 0) { b |= print_euler(graph[i][j], visited, count); } } // If the euler path is found if (b) { path.push_back(arr[i]); return true ; } // Else unmark the node else { visited[i] = 0; count--; return false ; } } // Function to create the graph and // print the required path void connect() { int n = arr.size(); graph.clear(); graph.resize(n); // Connect the nodes for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { if (i == j) continue ; // If the last character matches with the // first character if (arr[i][arr[i].length() - 1] == arr[j][0]) { graph[i].push_back(j); } } } // Print the path for ( int i = 0; i < n; i++) { int visited[n] = {0}, count = 0; // If the euler path starts // from the ith node if (print_euler(i, visited, count)) break ; } // Print the euler path for ( int i = path.size() - 1; i >= 0; i--) { cout << path[i]; if (i != 0) cout << " " ; } } // Driver code int main() { arr.push_back( "451" ); arr.push_back( "378" ); arr.push_back( "123" ); arr.push_back( "1254" ); // Create graph and print the path connect(); return 0; } |
Java
import java.util.ArrayList; public class EulerPath { // To store the array elements static ArrayList<String> arr = new ArrayList<>(); // Adjacency list for the graph nodes static ArrayList<ArrayList<Integer>> graph = new ArrayList<>(); // To store the Euler path static ArrayList<String> path = new ArrayList<>(); // Print Eulerian path static boolean printEuler( int i, int [] visited, int count) { // Mark node as visited and increase the count visited[i] = 1 ; count++; // If all the nodes are visited then we have traversed the Euler path if (count == graph.size()) { path.add(arr.get(i)); return true ; } // Check if the node lies in the Euler path boolean b = false ; // Traverse through remaining edges for ( int j = 0 ; j < graph.get(i).size(); j++) { if (visited[graph.get(i).get(j)] == 0 ) { b |= printEuler(graph.get(i).get(j), visited, count); } } // If the Euler path is found if (b) { path.add(arr.get(i)); return true ; } // Else unmark the node else { visited[i] = 0 ; count--; return false ; } } // Function to create the graph and print the required path static void connect() { int n = arr.size(); graph.clear(); // Initialize adjacency list for ( int i = 0 ; i < n; i++) { graph.add( new ArrayList<>()); } // Connect the nodes for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { if (i == j) continue ; // If the last character matches with the first character if (arr.get(i).charAt(arr.get(i).length() - 1 ) == arr.get(j).charAt( 0 )) { graph.get(i).add(j); } } } // Print the path for ( int i = 0 ; i < n; i++) { int [] visited = new int [n]; int count = 0 ; // If the Euler path starts from the ith node if (printEuler(i, visited, count)) { break ; } } // Print the Euler path for ( int i = path.size() - 1 ; i >= 0 ; i--) { System.out.print(path.get(i)); if (i != 0 ) System.out.print( " " ); } } // Driver code public static void main(String[] args) { arr.add( "451" ); arr.add( "378" ); arr.add( "123" ); arr.add( "1254" ); // Create graph and print the path connect(); } } |
Python3
# To store the array elements arr = [] # Adjacency list for the graph nodes graph = [] # To store the euler path path = [] # Print eulerian path def print_euler(i, visited, count): global graph, arr, path # Mark node as visited and increase the count visited[i] = 1 count + = 1 # If all the nodes are visited then we have traversed the euler path if count = = len (graph): path.append(arr[i]) return True # Check if the node lies in euler path b = False # Traverse through remaining edges for j in graph[i]: if visited[j] = = 0 : b | = print_euler(j, visited, count) # If the euler path is found if b: path.append(arr[i]) return True # Else unmark the node else : visited[i] = 0 count - = 1 return False # Function to create the graph and print the required path def connect(): global graph, arr, path n = len (arr) graph = [[] for _ in range (n)] # Connect the nodes for i in range (n): for j in range (n): if i = = j: continue # If the last character matches with the first character if arr[i][ - 1 ] = = arr[j][ 0 ]: graph[i].append(j) # Print the path for i in range (n): visited = [ 0 ] * n count = 0 # If the euler path starts from the ith node if print_euler(i, visited, count): break # Print the euler path print ( " " .join(path[:: - 1 ])) # Driver code if __name__ = = "__main__" : arr = [ "451" , "378" , "123" , "1254" ] # Create graph and print the path connect() |
C#
using System; using System.Collections.Generic; namespace EulerianPath { class GFG { // To store the array elements static List< string > arr = new List< string >(); // Adjacency list for the graph nodes static List<List< int >> graph = new List<List< int >>(); // To store the euler path static List< string > path = new List< string >(); // Print eulerian path static bool PrintEuler( int i, int [] visited, int count) { // Mark node as visited and increase the count visited[i] = 1; count++; // If all the nodes are visited, then we have traversed the euler path if (count == graph.Count) { path.Add(arr[i]); return true ; } // Check if the node lies in euler path bool b = false ; // Traverse through remaining edges foreach ( int j in graph[i]) { if (visited[j] == 0) { b |= PrintEuler(j, visited, count); } } // If the euler path is found if (b) { path.Add(arr[i]); return true ; } // Else unmark the node else { visited[i] = 0; count--; return false ; } } // Function to create the graph and print the required path static void Connect() { int n = arr.Count; graph.Clear(); for ( int i = 0; i < n; i++) { graph.Add( new List< int >()); } // Connect the nodes for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { if (i == j) continue ; // If the last character matches with the first character if (arr[i][arr[i].Length - 1] == arr[j][0]) { graph[i].Add(j); } } } // Print the path for ( int i = 0; i < n; i++) { int [] visited = new int [n]; int count = 0; // If the euler path starts from the ith node if (PrintEuler(i, visited, count)) break ; } // Print the euler path for ( int i = path.Count - 1; i >= 0; i--) { Console.Write(path[i]); if (i != 0) Console.Write( " " ); } } // Driver code static void Main( string [] args) { arr.Add( "451" ); arr.Add( "378" ); arr.Add( "123" ); arr.Add( "1254" ); // Create graph and print the path Connect(); } } } |
Javascript
// To store the array elements const arr = [ "451" , "378" , "123" , "1254" ]; // Adjacency list for the graph nodes const graph = []; // To store the euler path const path = []; // Print eulerian path function printEuler(i, visited, count) { // Mark node as visited and increase the count visited[i] = 1; count++; // If all the nodes are visited, then we have traversed the Euler path if (count === graph.length) { path.push(arr[i]); return true ; } // Check if the node lies in the Euler path let found = false ; // Traverse through remaining edges for (let j = 0; j < graph[i].length; j++) { if (visited[graph[i][j]] === 0) { found |= printEuler(graph[i][j], visited, count); } } // If the Euler path is found if (found) { path.push(arr[i]); return true ; } // Else unmark the node else { visited[i] = 0; count--; return false ; } } // Function to create the graph and print the required path function connect() { const n = arr.length; graph.length = 0; for (let i = 0; i < n; i++) { graph.push([]); for (let j = 0; j < n; j++) { if (i === j) continue ; // If the last character matches with the first character if (arr[i][arr[i].length - 1] === arr[j][0]) { graph[i].push(j); } } } // Print the path for (let i = 0; i < n; i++) { const visited = Array(n).fill(0); let count = 0; // If the Euler path starts from the ith node if (printEuler(i, visited, count)) break ; } // Print the Euler path for (let i = path.length - 1; i >= 0; i--) { process.stdout.write(path[i]); if (i !== 0) process.stdout.write( " " ); } } // Driver code connect(); |
1254 451 123 378
Time Complexity : O(n^2 * 2^n), where n is the number of strings in the input array. This is because for each pair of strings, we are comparing the last character of one string with the first character of the other string, and this takes O(n^2) time. In addition, we are using a recursive function to find the eulerian path, and the worst-case scenario involves traversing all the edges in the graph, which takes O(2^n) time.
Auxiliary Space:O(n + m), where n is the number of strings in the input array and m is the total number of edges in the graph. This is because we are using several auxiliary data structures such as vectors to store the array elements, adjacency list for the graph nodes, and the eulerian path. In addition, we are using an array to keep track of the visited nodes during the depth-first search, which takes O(n) space.