README_EN.md

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

Python3

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

Java

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];
    }
}

Go

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]
}

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