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 <= 10001 <= sentence1[i].length, sentence2[i].length <= 20sentence1[i] and sentence2[i] consist of lower-case and upper-case English letters.0 <= similarPairs.length <= 2000similarPairs[i].length == 21 <= xi.length, yi.length <= 20xi and yi consist of English letters.
Solutions#
Solution 1#
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| 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
|
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| 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];
}
}
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| 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];
}
};
|
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| 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]
}
|
