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]
andsentence2[i]
are similar.
Notice that a word is always similar to itself, also notice that the similarity relation is not transitive. For example, if the words a
and b
are similar, and the words b
and c
are similar, a
and c
are not necessarily similar.
Example 1:
Input: sentence1 = ["great","acting","skills"], sentence2 = ["fine","drama","talent"], similarPairs = [["great","fine"],["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 = ["great"], sentence2 = ["great"], similarPairs = [] Output: true Explanation: A word is similar to itself.
Example 3:
Input: sentence1 = ["great"], sentence2 = ["doubleplus","good"], similarPairs = [["great","doubleplus"]] Output: false Explanation: As they don't have the same length, we return false.
Constraints:
1 <= sentence1.length, sentence2.length <= 1000
1 <= sentence1[i].length, sentence2[i].length <= 20
sentence1[i]
andsentence2[i]
consist of English letters.0 <= similarPairs.length <= 1000
similarPairs[i].length == 2
1 <= xi.length, yi.length <= 20
xi
andyi
consist of lower-case and upper-case English letters.- All the pairs
(xi, yi)
are distinct.
class Solution:
def areSentencesSimilar(
self, sentence1: List[str], sentence2: List[str], similarPairs: List[List[str]]
) -> bool:
if len(sentence1) != len(sentence2):
return False
s = {(a, b) for a, b in similarPairs}
return all(
a == b or (a, b) in s or (b, a) in s for a, b in zip(sentence1, sentence2)
)
class Solution {
public boolean areSentencesSimilar(
String[] sentence1, String[] sentence2, List<List<String>> similarPairs) {
if (sentence1.length != sentence2.length) {
return false;
}
Set<String> s = new HashSet<>();
for (List<String> e : similarPairs) {
s.add(e.get(0) + "." + e.get(1));
}
for (int i = 0; i < sentence1.length; ++i) {
String a = sentence1[i], b = sentence2[i];
if (!a.equals(b) && !s.contains(a + "." + b) && !s.contains(b + "." + a)) {
return false;
}
}
return true;
}
}
class Solution {
public:
bool areSentencesSimilar(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) {
int m = sentence1.size(), n = sentence2.size();
if (m != n) return false;
unordered_set<string> s;
for (auto e : similarPairs) s.insert(e[0] + "." + e[1]);
for (int i = 0; i < n; ++i) {
string a = sentence1[i], b = sentence2[i];
if (a != b && !s.count(a + "." + b) && !s.count(b + "." + a)) return false;
}
return true;
}
};
func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs [][]string) bool {
if len(sentence1) != len(sentence2) {
return false
}
s := map[string]bool{}
for _, e := range similarPairs {
s[e[0]+"."+e[1]] = true
}
for i, a := range sentence1 {
b := sentence2[i]
if a != b && !s[a+"."+b] && !s[b+"."+a] {
return false
}
}
return true
}