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: 123 321



Input: 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++

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// 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;
}

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# 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 



















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Output:


1254 451 123 378








    

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