Given a sorted dictionary of an alien language, find order of characters

Given a sorted dictionary (array of words) of an alien language, find order of characters in the language.

 

Examples:

Input:  words[] = {"baa", "abcd", "abca", "cab", "cad"}
Output: Order of characters is 'b', 'd', 'a', 'c'
Note that words are sorted and in the given language "baa" 
comes before "abcd", therefore 'b' is before 'a' in output.
Similarly we can find other orders.

Input:  words[] = {"caa", "aaa", "aab"}
Output: Order of characters is 'c', 'a', 'b'

The idea is to create a graph of characters and then find topological sorting of the created graph. Following are the detailed steps.

1) Create a graph g with number of vertices equal to the size of alphabet in the given alien language. For example, if the alphabet size is 5, then there can be 5 characters in words. Initially there are no edges in graph.

2) Do following for every pair of adjacent words in given sorted array.
…..a) Let the current pair of words be word1 and word2. One by one compare characters of both words and find the first mismatching characters.
…..b) Create an edge in g from mismatching character of word1 to that of word2.

3) Print topological sorting of the above created graph.

 

// A C++ program to order of characters in an alien language
#include<iostream>
#include <list>
#include <stack>
#include <cstring>
using namespace std;
 
// Class to represent a graph
class Graph
{
    int V;    // No. of vertices'
 
    // Pointer to an array containing adjacency listsList
    list<int> *adj;
 
    // A function used by topologicalSort
    void topologicalSortUtil(int v, bool visited[], stack<int> &Stack);
public:
    Graph(int V);   // Constructor
 
    // function to add an edge to graph
    void addEdge(int v, int w);
 
    // prints a Topological Sort of the complete graph
    void topologicalSort();
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
 
void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); // Add w to v’s list.
}
 
// A recursive function used by topologicalSort
void Graph::topologicalSortUtil(int v, bool visited[], stack<int> &Stack)
{
    // Mark the current node as visited.
    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])
            topologicalSortUtil(*i, visited, Stack);
 
    // Push current vertex to stack which stores result
    Stack.push(v);
}
 
// The function to do Topological Sort. It uses recursive topologicalSortUtil()
void Graph::topologicalSort()
{
    stack<int> Stack;
 
    // Mark all the vertices as not visited
    bool *visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
    // Call the recursive helper function to store Topological Sort
    // starting from all vertices one by one
    for (int i = 0; i < V; i++)
        if (visited[i] == false)
            topologicalSortUtil(i, visited, Stack);
 
    // Print contents of stack
    while (Stack.empty() == false)
    {
        cout << (char) ('a' + Stack.top()) << " ";
        Stack.pop();
    }
}
 
int min(int x, int y)
{
    return (x < y)? x : y;
}
 
// This function fidns and prints order of characer from a sorted
// array of words. n is size of words[].  alpha is set of possible
// alphabets.
// For simplicity, this function is written in a way that only
// first 'alpha' characters can be there in words array.  For
// example if alpha is 7, then words[] should have only 'a', 'b',
// 'c' 'd', 'e', 'f', 'g'
void printOrder(string words[], int n, int alpha)
{
    // Create a graph with 'aplha' edges
    Graph g(alpha);
 
    // Process all adjacent pairs of words and create a graph
    for (int i = 0; i < n-1; i++)
    {
        // Take the current two words and find the first mismatching
        // character
        string word1 = words[i], word2 = words[i+1];
        for (int j = 0; j < min(word1.length(), word2.length()); j++)
        {
            // If we find a mismatching character, then add an edge
            // from character of word1 to that of word2
            if (word1[j] != word2[j])
            {
                g.addEdge(word1[j]-'a', word2[j]-'a');
                break;
            }
        }
    }
 
    // Print topological sort of the above created graph
    g.topologicalSort();
}
 
// Driver program to test above functions
int main()
{
    string words[] = {"caa", "aaa", "aab"};
    printOrder(words, 3, 3);
    return 0;
}

 Time Complexity: The first step to create a graph takes O(n + alhpa) time where n is number of given words and alpha is number of characters in given alphabet. The second step is also topological sorting. Note that there would be alpha vertices and at-most (n-1) edges in the graph. The time complexity of topological sorting is O(V+E) which is O(n + aplha) here. So overall time complexity is O(n + aplha) + O(n + aplha) which is O(n + aplha).

来自：http://www.geeksforgeeks.org/given-sorted-dictionary-find-precedence-characters/