LeetCode: Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:

Given the following words in dictionary,

[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

The correct order is: "wertf".

Example 2:

Given the following words in dictionary,

[
  "z",
  "x"
]

The correct order is: "zx".

Example 3:

Given the following words in dictionary,

[
  "z",
  "x",
  "z"
]

The order is invalid, so return "".

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

public class Solution {
        public String alienOrder(String[] words) {
        if (words == null || words.length == 0) {
            return "";
        }

        if (words.length == 1) {
            return words[0];
        }

        Map<Character, Set<Character>> graph = new HashMap<>();
        Map<Character, Integer> toCount = new HashMap<>();
        StringBuilder sb = new StringBuilder();

        init(graph, toCount, words);
        process(graph, toCount, words);
        build(sb, graph, toCount);
         return sb.length()== toCount.size()? sb.toString():"";
    }

    private void init(Map<Character, Set<Character>> graph, Map<Character, Integer> toCount, String[] words) {
        Arrays.stream(words).forEach(word -> {
            for (char c : word.toCharArray()) {
                graph.putIfAbsent(c, new HashSet<>());
                toCount.putIfAbsent(c, 0);
            }
        });
    }

    private void process(Map<Character, Set<Character>> graph, Map<Character, Integer> toCount, String[] words) {
        Set<String> edges = new HashSet<>();
        for (int i = 0; i < words.length - 1; i++) {
            String s1 = words[i];
            String s2 = words[i + 1];

            // TODO: what if s1 or s2 is null
            for (int j = 0; j < s1.length() && j < s2.length(); j++) {
                char c1 = s1.charAt(j);
                char c2 = s2.charAt(j);
                if (c1 == c2) {
                    continue;
                }
                String edge = "" + c1 + c2;
                if (!edges.contains(edge)) {
                    toCount.put(c2, toCount.get(c2) + 1);
                    Set<Character> val = graph.get(c1);
                    val.add(c2);
                    graph.put(c1, val);
                    edges.add(edge);
                    break;
                }
            }
        }
    }

    private void build(StringBuilder sb, Map<Character, Set<Character>> graph, Map<Character, Integer> toCount) {
        Queue<Character> queue = new LinkedList<>();
        toCount.forEach((key, value) -> {
            if (value == 0) {
                queue.add(key);
            }
        });

        while (!queue.isEmpty()) {
            char c = queue.poll();
            sb.append(c);
            graph.get(c).forEach(ch -> {
                int value = toCount.get(ch) - 1;
                toCount.put(ch, value);
                if (value == 0) {
                    queue.add(ch);
                }
            });
        }
    }
}