LeetCode 0734. Sentence Similarity Solution in Java, C++, Python & More | Explanation + Code

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 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] and sentence2[i] consist of English letters.
• 0 <= similarPairs.length <= 1000
• similarPairs[i].length == 2
• 1 <= xi.length, yi.length <= 20
• xi and yi consist of lower-case and upper-case English letters.
• All the pairs (xi, yi) are distinct.

Solutions

Solution 1: Hash Table

First, we check if the lengths of sentence1 and sentence2 are equal. If they are not equal, return false.

Then we use a hash table s to store all similar word pairs. For each word pair [x, y] in similarPairs, we add x and y to the hash table s.

Next, we traverse sentence1 and sentence2. For each position i, if sentence1[i] is not equal to sentence2[i], and (sentence1[i], sentence2[i]) and (sentence2[i], sentence1[i]) are not in the hash table s, then return false.

If the traversal ends without returning false, it means sentence1 and sentence2 are similar, so return true.

The time complexity is O(L), and the space complexity is O(L), where L is the sum of the lengths of all strings in the problem.

PythonJavaC++GoTypeScriptRustJavaScript

class Solution: def areSentencesSimilar( self, sentence1: List[str], sentence2: List[str], similarPairs: List[List[str]] ) -> bool: if len(sentence1) != len(sentence2): return False s = {(x, y) for x, y in similarPairs} for x, y in zip(sentence1, sentence2): if x != y and (x, y) not in s and (y, x) not in s: return False return True(code-box)

class Solution { public boolean areSentencesSimilar( String[] sentence1, String[] sentence2, List<List<String>> similarPairs) { if (sentence1.length != sentence2.length) { return false; } Set<List<String>> s = new HashSet<>(); for (var p : similarPairs) { s.add(p); } for (int i = 0; i < sentence1.length; i++) { if (!sentence1[i].equals(sentence2[i]) && !s.contains(List.of(sentence1[i], sentence2[i])) && !s.contains(List.of(sentence2[i], sentence1[i]))) { return false; } } return true; } }(code-box)

class Solution { public: bool areSentencesSimilar(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) { if (sentence1.size() != sentence2.size()) { return false; } unordered_set<string> s; for (const auto& p : similarPairs) { s.insert(p[0] + "#" + p[1]); s.insert(p[1] + "#" + p[0]); } for (int i = 0; i < sentence1.size(); ++i) { if (sentence1[i] != sentence2[i] && !s.contains(sentence1[i] + "#" + sentence2[i])) { return false; } } return true; } };(code-box)

func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs [][]string) bool { if len(sentence1) != len(sentence2) { return false } s := map[string]bool{} for _, p := range similarPairs { s[p[0]+"#"+p[1]] = true } for i, x := range sentence1 { y := sentence2[i] if x != y && !s[x+"#"+y] && !s[y+"#"+x] { return false } } return true }(code-box)

function areSentencesSimilar( sentence1: string[], sentence2: string[], similarPairs: string[][], ): boolean { if (sentence1.length !== sentence2.length) { return false; } const s = new Set<string>(); for (const [x, y] of similarPairs) { s.add(x + '#' + y); s.add(y + '#' + x); } for (let i = 0; i < sentence1.length; i++) { if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) { return false; } } return true; }(code-box)

use std::collections::HashSet; impl Solution { pub fn are_sentences_similar( sentence1: Vec<String>, sentence2: Vec<String>, similar_pairs: Vec<Vec<String>>, ) -> bool { if sentence1.len() != sentence2.len() { return false; } let s: HashSet<(String, String)> = similar_pairs .into_iter() .map(|pair| (pair[0].clone(), pair[1].clone())) .collect(); for (x, y) in sentence1.iter().zip(sentence2.iter()) { if x != y && !s.contains(&(x.clone(), y.clone())) && !s.contains(&(y.clone(), x.clone())) { return false; } } true } }(code-box)

/** * @param {string[]} sentence1 * @param {string[]} sentence2 * @param {string[][]} similarPairs * @return {boolean} */ var areSentencesSimilar = function (sentence1, sentence2, similarPairs) { if (sentence1.length !== sentence2.length) { return false; } const s = new Set(); for (const [x, y] of similarPairs) { s.add(x + '#' + y); s.add(y + '#' + x); } for (let i = 0; i < sentence1.length; i++) { if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) { return false; } } return true; };(code-box)