Leetcode #737: Sentence Similarity II
In this guide, we solve Leetcode #737 Sentence Similarity II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search, Union Find, Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.