How do you solve Short Encoding of Words on LeetCode?

Short Encoding of Words (LeetCode #820) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Understanding suffix relationships between words can help minimize the reference string length.

What is the brute force solution for Short Encoding of Words?

The brute force approach for Short Encoding of Words is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a reference string by concatenating each word followed by a '#' character. This method checks all possible combinations of words to ensure no overlaps, leading to a longer string. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Short Encoding of Words?

Short Encoding of Words uses the following concepts: Array, Hash Table, String, Trie. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Short Encoding of Words?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as single-word inputs or completely overlapping words. Practice explaining your thought process out loud as you code to simulate the interview environment.

What are common mistakes when solving Short Encoding of Words?

Not considering overlapping words, leading to incorrect total lengths. Failing to account for the '#' character when calculating lengths.

#820

Short Encoding of Words

Medium

ArrayHash TableStringTrieHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

💡

Intuition

Time O(n * m)Space O(n)

The optimal approach leverages the fact that we can avoid duplicating parts of words by only encoding the unique suffixes. By storing the words in a set, we can efficiently check for overlaps and minimize the reference string length.

⚙️

Algorithm

4 steps

1Step 1: Create a set to store all words.
2Step 2: For each word, check if it is a suffix of any other word in the set.
3Step 3: If not, add the length of the word plus one (for the '#') to the total length.
4Step 4: Return the total length.

solution.py8 lines

1# Full working Python code
2words = ['time', 'me', 'bell']
3word_set = set(words)
4total_length = 0
5for word in words:
6    if not any(word != other and other.endswith(word) for other in word_set):
7        total_length += len(word) + 1
8print(total_length)

ℹ

Complexity note: This complexity arises because we check each word against all others, where `m` is the average length of the words. The space complexity is due to storing the words in a set.

1Understanding suffix relationships between words can help minimize the reference string length.
2Using data structures like sets can improve lookup times and help manage word uniqueness.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.