LeetCode 269: Alien Dictionary

Posted 2020-10-05 keepshuatishuati

Notes: Lots of places need to be remember:

Edge cases:

　　1. Only one word, still need to be calculated as it represents.

　　2. All unique chars that not be placed in the result means there are several cycles. It must return empty string.

Data Structure:

　　1. Better use HashSet in Map. Otherwise, it could be duplicates.

　　2. Remove keys from map should be outside of keySet loop.

 

class Solution {
    private Map<Character, Set<Character>> dict;
    public String alienOrder(String[] words) {
        if (words.length == 0) {
            return "";
        }
        
        dict = new HashMap<>();
        for (String word : words) {
            for (char c : word.toCharArray()) {
                if (!dict.containsKey(c)) {
                    dict.put(c, new HashSet<>());
                }
            }
        }
        int total = dict.size();
        
        for (int i = 0; i < words.length - 1; i++) {
            generateDependencies(words[i], words[i + 1]);
        }

        Queue<Character> queue = new LinkedList<>();
        List<Character> toRemove = new ArrayList<>();
        for (char c : dict.keySet()) {
            if (dict.get(c).size() == 0) {
                queue.offer(c);
                toRemove.add(c);
            }
        }
        dict.keySet().removeAll(toRemove);

        StringBuilder result = new StringBuilder();
        while (!queue.isEmpty()) {
            int length = queue.size();
            for (int i = 0; i < length; i++) {
                char current = queue.poll();
                toRemove.clear();
                for (char c : dict.keySet()) {
                    if (dict.get(c).contains(current)) {
                        dict.get(c).remove(new Character(current));
                    }
                    if (dict.get(c).size() == 0) {
                        queue.offer(c);
                        toRemove.add(c);
                    }
                }
                dict.keySet().removeAll(toRemove);
                result.append(current);
            }
        }
        return total == result.length() ? result.toString() : "";
    }
    
    private void generateDependencies(String s1, String s2) {
        int index = 0;
        while (index < s1.length() && index < s2.length()) {
            if (s1.charAt(index) != s2.charAt(index)) {
                dict.get(s2.charAt(index)).add(s1.charAt(index));
                break;
            }
            index++;
        }
    }
}