Question

737. Sentence Similarity II

中文文档

Description

We can represent a sentence as an array of words, for example, the sentence "I am happy with leetcode" can be represented as arr = ["I","am",happy","with","leetcode"].

Given two sentences sentence1 and sentence2 each represented as a string array and given an array of string pairs similarPairs where similarPairs[i] = [xi, yi] indicates that the two words xi and yi are similar.

Return true_ if sentence1 and sentence2 are similar, or _false_ if they are not similar_.

Two sentences are similar if:

• They have the same length (i.e., the same number of words)
• sentence1[i] and sentence2[i] are similar.

Notice that a word is always similar to itself, also notice that the similarity relation is transitive. For example, if the words a and b are similar, and the words b and c are similar, then a and c are similar.

 

Example 1:

Input: sentence1 = ["great","acting","skills"], sentence2 = ["fine","drama","talent"], similarPairs = [["great","good"],["fine","good"],["drama","acting"],["skills","talent"]]
Output: true
Explanation: The two sentences have the same length and each word i of sentence1 is also similar to the corresponding word in sentence2.

Example 2:

Input: sentence1 = ["I","love","leetcode"], sentence2 = ["I","love","onepiece"], similarPairs = [["manga","onepiece"],["platform","anime"],["leetcode","platform"],["anime","manga"]]
Output: true
Explanation: "leetcode" --> "platform" --> "anime" --> "manga" --> "onepiece".
Since "leetcode is similar to "onepiece" and the first two words are the same, the two sentences are similar.

Example 3:

Input: sentence1 = ["I","love","leetcode"], sentence2 = ["I","love","onepiece"], similarPairs = [["manga","hunterXhunter"],["platform","anime"],["leetcode","platform"],["anime","manga"]]
Output: false
Explanation: "leetcode" is not similar to "onepiece".

 

Constraints:

• 1 <= sentence1.length, sentence2.length <= 1000
• 1 <= sentence1[i].length, sentence2[i].length <= 20
• sentence1[i] and sentence2[i] consist of lower-case and upper-case English letters.
• 0 <= similarPairs.length <= 2000
• similarPairs[i].length == 2
• 1 <= xi.length, yi.length <= 20
• xi and yi consist of English letters.

Solutions

Solution 1

class Solution:
  def areSentencesSimilarTwo(
    self, sentence1: List[str], sentence2: List[str], similarPairs: List[List[str]]
  ) -> bool:
    if len(sentence1) != len(sentence2):
      return False
    n = len(similarPairs)
    p = list(range(n << 1))

    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    words = {}
    idx = 0
    for a, b in similarPairs:
      if a not in words:
        words[a] = idx
        idx += 1
      if b not in words:
        words[b] = idx
        idx += 1
      p[find(words[a])] = find(words[b])

    for i in range(len(sentence1)):
      if sentence1[i] == sentence2[i]:
        continue
      if (
        sentence1[i] not in words
        or sentence2[i] not in words
        or find(words[sentence1[i]]) != find(words[sentence2[i]])
      ):
        return False
    return True

class Solution {
  private int[] p;

  public boolean areSentencesSimilarTwo(
    String[] sentence1, String[] sentence2, List<List<String>> similarPairs) {
    if (sentence1.length != sentence2.length) {
      return false;
    }
    int n = similarPairs.size();
    p = new int[n << 1];
    for (int i = 0; i < p.length; ++i) {
      p[i] = i;
    }
    Map<String, Integer> words = new HashMap<>();
    int idx = 0;
    for (List<String> e : similarPairs) {
      String a = e.get(0), b = e.get(1);
      if (!words.containsKey(a)) {
        words.put(a, idx++);
      }
      if (!words.containsKey(b)) {
        words.put(b, idx++);
      }
      p[find(words.get(a))] = find(words.get(b));
    }
    for (int i = 0; i < sentence1.length; ++i) {
      if (Objects.equals(sentence1[i], sentence2[i])) {
        continue;
      }
      if (!words.containsKey(sentence1[i]) || !words.containsKey(sentence2[i])
        || find(words.get(sentence1[i])) != find(words.get(sentence2[i]))) {
        return false;
      }
    }
    return true;
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  vector<int> p;
  bool areSentencesSimilarTwo(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) {
    if (sentence1.size() != sentence2.size())
      return false;
    int n = similarPairs.size();
    p.resize(n << 1);
    for (int i = 0; i < p.size(); ++i)
      p[i] = i;
    unordered_map<string, int> words;
    int idx = 0;
    for (auto e : similarPairs) {
      string a = e[0], b = e[1];
      if (!words.count(a))
        words[a] = idx++;
      if (!words.count(b))
        words[b] = idx++;
      p[find(words[a])] = find(words[b]);
    }
    for (int i = 0; i < sentence1.size(); ++i) {
      if (sentence1[i] == sentence2[i])
        continue;
      if (!words.count(sentence1[i]) || !words.count(sentence2[i]) || find(words[sentence1[i]]) != find(words[sentence2[i]]))
        return false;
    }
    return true;
  }

  int find(int x) {
    if (p[x] != x)
      p[x] = find(p[x]);
    return p[x];
  }
};

var p []int

func areSentencesSimilarTwo(sentence1 []string, sentence2 []string, similarPairs [][]string) bool {
  if len(sentence1) != len(sentence2) {
    return false
  }
  n := len(similarPairs)
  p = make([]int, (n<<1)+10)
  for i := 0; i < len(p); i++ {
    p[i] = i
  }
  words := make(map[string]int)
  idx := 1
  for _, e := range similarPairs {
    a, b := e[0], e[1]
    if words[a] == 0 {
      words[a] = idx
      idx++
    }
    if words[b] == 0 {
      words[b] = idx
      idx++
    }
    p[find(words[a])] = find(words[b])
  }
  for i := 0; i < len(sentence1); i++ {
    if sentence1[i] == sentence2[i] {
      continue
    }
    if words[sentence1[i]] == 0 || words[sentence2[i]] == 0 || find(words[sentence1[i]]) != find(words[sentence2[i]]) {
      return false
    }
  }
  return true
}

func find(x int) int {
  if p[x] != x {
    p[x] = find(p[x])
  }
  return p[x]
}