LeetCode 527: Word Abbreviation Solution in Python – A Step-by-Step Guide

Imagine you’re tasked with shortening a list of words—like “internationalization,” “localization,” and “like”—into unique abbreviations, such as “i18n,” “l10n,” and “like,” by keeping some letters and replacing stretches with their count. That’s the clever challenge of LeetCode 527: Word Abbreviation, a hard-level problem that’s a fantastic way to practice string manipulation in Python. We’ll explore two solutions: the Best Solution, a greedy approach with prefix grouping that’s efficient and smart, and an Alternative Solution, a brute-force abbreviation generation method that’s thorough but slower. With detailed examples, clear code, and a friendly tone—especially for the greedy strategy—this guide will help you abbreviate those words, whether you’re new to coding or leveling up. Let’s shorten some strings and start counting!

What Is LeetCode 527: Word Abbreviation?

In LeetCode 527: Word Abbreviation, you’re given a list of strings words, and your task is to return a list of abbreviations where each abbreviation is unique and formed by keeping the first character, replacing a middle section with its length, and keeping the last character (e.g., “internationalization” → “i18n”). The abbreviation must be shorter than the original word and distinguishable from others. For example, with words = ["like", "god", "internal", "me"], one valid output is ["l2e", "g2d", "i6l", "me"]. This problem builds on LeetCode 408: Valid Word Abbreviation.

Problem Statement

Input: List of strings words.
Output: List of strings—unique abbreviations for each word.
Rules: Abbreviation format: first char + count + last char; must be shorter than original; unique in list.

Constraints

1 <= words.length <= 20
4 <= words[i].length <= 20
words[i] contains only lowercase English letters.
All words are unique.

Examples

Input: ["like","god","internal","me"]

Output: ["l2e","g2d","i6l","me"]
Each abbreviation is unique and shorter.

Input: ["internationalization","localization"]

Output: ["i18n","l10n"]
Standard abbreviations, unique.

Input: ["abcd"]

Output: ["a2d"]
Single word abbreviated.

Understanding the Problem: Crafting Unique Abbreviations

To solve LeetCode 527: Word Abbreviation in Python, we need to generate abbreviations for each word in words that are shorter than the original and unique across the list, using the format “first + count + last.” A naive approach might try all possible abbreviations, but with constraints allowing efficient grouping, we can optimize. We’ll explore:

Best Solution (Greedy with Prefix Grouping): O(n * m) time, O(n) space (n = number of words, m = avg word length)—efficient and clever.
Alternative Solution (Brute-Force Abbreviation Generation): O(n m²) time, O(n m)—thorough but slow.

Let’s dive into the greedy solution with a friendly breakdown!

Best Solution: Greedy with Prefix Grouping

Why Greedy with Prefix Grouping Wins

The greedy with prefix grouping solution is the best for LeetCode 527 because it efficiently assigns abbreviations by grouping words with the same initial abbreviation, then incrementally increasing prefix length to resolve duplicates, running in O(n * m) time and O(n) space. It’s like sorting mail by zip code and tweaking addresses only when needed!

How It Works

Think of this as an abbreviation assembly line:

Step 1: Initial Abbreviation:

For each word, start with minimal abbreviation (e.g., “like” → “l2e”).

Step 2: Group by Abbreviation:

Use a dictionary to group words by their initial abbreviation.

Step 3: Resolve Duplicates:

For groups with >1 word, increase prefix length by 1, recompute abbreviations, repeat until unique.

Step 4: Return Results:

List of unique abbreviations in original order.

Why It Works:

Greedy prefix increase ensures uniqueness with minimal effort.
Grouping avoids unnecessary checks.

It’s like an abbreviation fine-tuner!

Step-by-Step Example

Example: words = ["like", "lake", "live"]

Step 1: Initial abbreviations:

“like” → “l2e”.
“lake” → “l2e”.
“live” → “l2e”.

Step 2: Group:

{"l2e": ["like", "lake", "live"]}.

Step 3: Resolve duplicates (prefix = 1):

“like” → “li2e”.
“lake” → “la2e”.
“live” → “li2e”.
New group: {"li2e": ["like", "live"], "la2e": ["lake"]}.
“li2e” still duplicates, increase prefix to 2:

“like” → “lik1e”.
“live” → “liv1e”.

Final: {"lik1e": ["like"], "liv1e": ["live"], "la2e": ["lake"]}.

Result: ["lik1e", "la2e", "liv1e"].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

class Solution:
    def wordsAbbreviation(self, words: List[str]) -> List[str]:
        def abbreviate(word, prefix_len):
            # Form abbreviation: prefix + count + last char
            if len(word) - prefix_len <= 3:  # Too short to abbreviate
                return word
            count = len(word) - prefix_len - 1
            return word[:prefix_len] + str(count) + word[-1]

        n = len(words)
        # Step 1: Initial abbreviations
        abbrevs = [abbreviate(word, 1) for word in words]
        result = words.copy()

        # Step 2: Process duplicates
        for i in range(n):
            while True:
                # Group by current abbreviation
                abbrev_to_indices = {}
                for j in range(n):
                    if abbrevs[j] not in abbrev_to_indices:
                        abbrev_to_indices[abbrevs[j]] = []
                    abbrev_to_indices[abbrevs[j]].append(j)

                # Check if i's abbreviation is unique
                if len(abbrev_to_indices[abbrevs[i]]) == 1:
                    break

                # Step 3: Increase prefix for duplicates
                dup_indices = abbrev_to_indices[abbrevs[i]]
                for idx in dup_indices:
                    prefix_len = len(result[idx].split(str(len(result[idx]) - 2))[0])
                    abbrevs[idx] = abbreviate(words[idx], prefix_len + 1)
                    result[idx] = abbrevs[idx]

        # Step 4: Return final abbreviations
        return result

Lines 3-8: abbreviate: Create abbreviation with prefix length.
Lines 10-12: Initial setup with prefix = 1.
Lines 15-29: For each word:

Group by abbreviation, resolve duplicates by increasing prefix.

Line 32: Return result.
Time Complexity: O(n * m)—n words, m avg length for prefix adjustments.
Space Complexity: O(n)—grouping dictionary.

It’s like an abbreviation optimizer!

Alternative Solution: Brute-Force Abbreviation Generation

Why an Alternative Approach?

The brute-force solution generates all possible abbreviations for each word and selects the shortest unique ones. It’s O(n * m²) time and O(n * m) space—thorough but inefficient due to excessive generation. Great for understanding all possibilities!

How It Works

Picture this as an abbreviation factory:

Step 1: Generate all abbreviations per word (varying prefix length).
Step 2: Check uniqueness against others.
Step 3: Pick shortest unique abbreviation.

It’s like an abbreviation overproducer!

Step-by-Step Example

Example: words = ["like", "lake"]

Step 1: Generate:

“like”: “l2e”, “li2e”, “lik1e”.
“lake”: “l2e”, “la2e”, “lak1e”.

Step 2: Check:

“l2e” duplicates, skip.
“li2e” unique for “like”, “la2e” unique for “lake”.

Result: ["li2e", "la2e"].

Code for Brute-Force Approach

class Solution:
    def wordsAbbreviation(self, words: List[str]) -> List[str]:
        def get_abbrevs(word):
            abbrevs = []
            for prefix_len in range(1, len(word) - 2):
                count = len(word) - prefix_len - 1
                abbrevs.append(word[:prefix_len] + str(count) + word[-1])
            return abbrevs

        n = len(words)
        all_abbrevs = [get_abbrevs(word) for word in words]
        result = [None] * n

        for i in range(n):
            for abbrev in all_abbrevs[i]:
                unique = True
                for j in range(n):
                    if i != j and abbrev in all_abbrevs[j]:
                        unique = False
                        break
                if unique:
                    result[i] = abbrev
                    break

        return result

Lines 3-7: Generate all abbreviations.
Lines 10-19: Check each abbreviation for uniqueness.
Time Complexity: O(n * m²)—generate and check.
Space Complexity: O(n * m)—store abbreviations.

It’s a brute-force abbreviator!

Comparing the Two Solutions

Greedy (Best):

Pros: O(n * m), O(n), efficient.
Cons: Grouping logic.

Brute-Force (Alternative):

Pros: O(n * m²), explicit.
Cons: Slower, more space.

Greedy wins for efficiency!

Additional Examples and Edge Cases

["abcd"]: ["a2d"].
["aaaa", "aaab"]: ["aa2a", "aa2b"].
["word"]: ["w2d"].

Greedy handles them all!

Complexity Recap

Greedy: Time O(n * m), Space O(n).
Brute-Force: Time O(n m²), Space O(n m).

Greedy’s the speed champ!

Key Takeaways

Greedy: Prefix tuning—learn at Python Basics!
Brute Force: Full but slow.
Strings: Abbreviation fun.
Python: Dicts and loops shine!

Final Thoughts: Shorten Those Words!

LeetCode 527: Word Abbreviation in Python is a clever string challenge. Greedy with prefix grouping is your fast track, while brute force shows all options. Want more? Try LeetCode 408: Valid Word Abbreviation or LeetCode 320: Generalized Abbreviation. Ready to abbreviate? Head to Solve LeetCode 527 on LeetCode and shorten those words today!