Question

link

Given set of words that are lexicographically sorted, return lexicographic order. E.g:

abc
acd
bcc
bed
bdc
dab

The order of letters for the given example would be

a->b->c->e->d

Solution

Just form a graph(DAG) and do a topological sort.

Start from the first pair in the dictionary. Compare two strings in this pair till first mismatch.

Eg: aad & aab, in this case create an edge d -> b.

More details is available here.

Variant, and a different solution

Another way of asking this question, is:

有一个字典因为某种原因每个字符都被替换成一个别的字符了（但还是一一对应)， 但是单词的顺序没有改变，比如

cat
coffee
common

--> 

dkc
dbhhzz
dbllbq

让找出的这个替换的规则（guaranteed to have a unique one)

Alternative solution is to check according to frequencies.

Code

public String lexicoOrder(String[] dict) {
    String ans = "";

    // for each pair, maintain 2 HashMap
    HashMap<Character, Integer> incount = new HashMap<Character, Integer>();
    HashMap<Character, List<Character>> connection = new HashMap<Character, List<Character>>();
    for (String str : dict) {
        for (char c : str.toCharArray()) {
            incount.put(c, 0);
            connection.put(c, new ArrayList<Character>());
        }
    }
    buildGraph(dict, incount, connection);

    // start topology sorting
    Queue<Character> temp = new LinkedList<Character>();
    for (char c : incount.keySet()) {
        if (incount.get(c) == 0) {
            temp.offer(c);
            incount.remove(c);
            // remove any node whose incount is 0
        }
    }
    while (!temp.isEmpty()) {
        char top = temp.poll();
        ans += top;
        // 'top' is next char in line. remove it and delete connections
        List<Character> inNodes = connection.get(top);
        for (char c : inNodes) {
            // remove incount for all nodes from inNodes
            incount.put(c, incount.get(c) - 1);
            if (incount.get(c) == 0) {
                incount.remove(c);
                temp.offer(c);
            }
        }
    }
    if (incount.size() == 0)
        return ans;
    else
        return "unable to find a valid char sequence.";
}

public void buildGraph(String[] dict, HashMap<Character, Integer> incount,
        HashMap<Character, List<Character>> connection) {
    // build the graph map
    // abc
    // acd
    // bcc
    // bed
    // bdc
    // dab
    for (int i = 0; i < dict.length - 1; i++) {
        // compare dict[i] and dict[i+1]
        String str1 = dict[i];
        String str2 = dict[i + 1];
        int p = 0;
        while (p < str1.length() && p < str2.length()) {
            if (str1.charAt(p) == str2.charAt(p)) {
                p++;
            } else {
                break;
            }
        }
        if (p == str1.length()) {
            // this is special case eg. "ab" & "abc"
            // this will not give up any information about lexico order
            continue;
        }
        char from = str1.charAt(p);
        char to = str2.charAt(p);
        // update incount
        incount.put(to, incount.get(to) + 1);
        // update connection
        connection.get(from).add(to);
    }
}