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@forkercat
Created November 15, 2019 08:32
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Alien Dictionary
class Solution {
public String alienOrder(String[] words) {
if (words == null || words.length == 0) {
return "";
}
Map<Character, Set<Character>> graph = new HashMap<>(); // List -> Set?
Map<Character, Integer> indegree = new HashMap<>();
buildGraph(words, graph, indegree);
// bfs
Queue<Character> queue = new LinkedList<>();
// push all 0-indegree nodes
for (char v : graph.keySet()) {
if (indegree.get(v) == 0) {
queue.offer(v);
}
}
int count = graph.size(); // number of nodes in total
StringBuilder sb = new StringBuilder();
while (queue.size() > 0) {
char v = queue.poll();
sb.append(v);
--count;
Set<Character> neighborList = graph.get(v);
// for each neighbor w
for (char w : neighborList) {
int degree = indegree.get(w);
indegree.put(w, degree - 1);
if (degree - 1 == 0) {
queue.offer(w);
}
}
}
if (count == 0) return sb.toString();
else return "";
}
private void buildGraph(String[] words, Map<Character, Set<Character>> graph, Map<Character, Integer> indegree) {
// preprocess
for (int i = 0; i < words.length; ++i) {
for (int j = 0; j < words[i].length(); ++j) {
char ch = words[i].charAt(j);
graph.put(ch, graph.getOrDefault(ch, new HashSet<>()));
indegree.put(ch, 0);
}
}
// add edges and calculate indegrees
for (int i = 0; i < words.length - 1; ++i) {
String s1 = words[i];
String s2 = words[i + 1];
int j = 0;
while (j < s1.length() && j < s2.length() && s1.charAt(j) == s2.charAt(j)) {
++j;
}
// consider: "abc" vs. "abcd", "abc" vs. "abc", "abcd" vs. "abc" -> no info. gained
if (j < s1.length() && j < s2.length()) {
char c1 = s1.charAt(j);
char c2 = s2.charAt(j);
if (graph.get(c1).contains(c2) == false) { // critical
graph.get(c1).add(c2);
indegree.put(c2, indegree.getOrDefault(c2, 0) + 1); // indegree
}
}
}
}
}
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