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SentenceSimilarityII737.java
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SentenceSimilarityII737.java
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/**
* Given two sentences words1, words2 (each represented as an array of strings),
* and a list of similar word pairs pairs, determine if two sentences are similar.
*
* For example, words1 = ["great", "acting", "skills"] and
* words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs
* are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"],
* ["skills","talent"]].
*
* Note that the similarity relation is transitive. For example, if "great"
* and "good" are similar, and "fine" and "good" are similar, then "great" and
* "fine" are similar.
*
* Similarity is also symmetric. For example, "great" and "fine" being similar
* is the same as "fine" and "great" being similar.
*
* Also, a word is always similar with itself. For example, the sentences
* words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though
* there are no specified similar word pairs.
*
* Finally, sentences can only be similar if they have the same number of words.
* So a sentence like words1 = ["great"] can never be similar to
* words2 = ["doubleplus","good"].
*
* Note:
* The length of words1 and words2 will not exceed 1000.
* The length of pairs will not exceed 2000.
* The length of each pairs[i] will be 2.
* The length of each words[i] and pairs[i][j] will be in the range [1, 20].
*/
public class SentenceSimilarityII737 {
public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) {
if (words1.length != words2.length) return false;
DisjointSet djs = new DisjointSet(pairs);
int len = words1.length;
for (int i=0; i<len; i++) {
String w1 = words1[i];
String w2 = words2[i];
if (w1.equals(w2)) continue;
String p1 = djs.find(w1);
String p2 = djs.find(w2);
if (p1 == null || p2 == null || !p1.equals(p2)) return false;
}
return true;
}
class DisjointSet {
Map<String, String> parent = new HashMap<>();
Map<String, Integer> rank = new HashMap<>();
public DisjointSet(String[][] pairs) {
for (String[] pair: pairs) {
if (!parent.containsKey(pair[0])) {
parent.put(pair[0], pair[0]);
rank.put(pair[0], 0);
}
if (!parent.containsKey(pair[1])) {
parent.put(pair[1], pair[1]);
rank.put(pair[1], 0);
}
union(pair[0], pair[1]);
}
}
public String find(String word) {
if (parent.containsKey(word) && !parent.get(word).equals(word)) {
parent.put(word, find(parent.get(word)));
}
return parent.get(word);
}
public void union(String w1, String w2) {
String p1 = find(w1);
String p2 = find(w2);
int r1 = rank.get(p1);
int r2 = rank.get(p2);
if (r1 > r2) {
parent.put(p2, p1);
} else if (r1 < r2) {
parent.put(p1, p2);
} else {
parent.put(p1, p2);
rank.put(p2, r2+1);
}
}
}
/**
* https://leetcode.com/problems/sentence-similarity-ii/solution/
*/
public boolean areSentencesSimilarTwo2(String[] words1, String[] words2, String[][] pairs) {
if (words1.length != words2.length) return false;
Map<String, Integer> index = new HashMap();
int count = 0;
DSU dsu = new DSU(2 * pairs.length);
for (String[] pair: pairs) {
for (String p: pair) if (!index.containsKey(p)) {
index.put(p, count++);
}
dsu.union(index.get(pair[0]), index.get(pair[1]));
}
for (int i = 0; i < words1.length; ++i) {
String w1 = words1[i], w2 = words2[i];
if (w1.equals(w2)) continue;
if (!index.containsKey(w1) || !index.containsKey(w2) ||
dsu.find(index.get(w1)) != dsu.find(index.get(w2)))
return false;
}
return true;
}
class DSU {
int[] parent;
public DSU(int N) {
parent = new int[N];
for (int i = 0; i < N; ++i)
parent[i] = i;
}
public int find(int x) {
if (parent[x] != x) parent[x] = find(parent[x]);
return parent[x];
}
public void union(int x, int y) {
parent[find(x)] = find(y);
}
}
}