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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// To store the array elementsvector<string> arr;// Adjacency list for the graph nodesvector<vector<int> > graph;// To store the euler pathvector<string> path;// Print eulerian pathbool 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 pathvoid 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 codeint 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;} |
Python3
# Python3 implementation of the approach# Print eulerian pathdef 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 pathdef 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 codeif __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 |
1254 451 123 378
Time Complexity : O(N* log(N))
Auxiliary Space: O(N)
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