Minimum Unique Word Abbreviation

Hard

Google

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Topics:

StringsBit Manipulation

You are given a target word and a dictionary of words. Your goal is to find the shortest unique abbreviation of the target word such that it does not conflict with any other word in the dictionary.

Abbreviation: An abbreviation can be formed by replacing any non-overlapping substring with the count of characters in that substring. For example, "apple" can be abbreviated as:
- "apple" (no abbreviation)
- "a4" (replace "pple" with 4)
- "4e" (replace "appl" with 4)
- "ap3" (replace "ple" with 3)
- "5" (replace "apple" with 5)
- "a3e" (replace "ppl" with 3)
Conflict: An abbreviation conflicts with a word in the dictionary if the word can be abbreviated to the same string.

For instance, if target = "apple" and dict = ["blade"], a valid abbreviation could be "a4" because "blade" cannot be abbreviated to "a4". However, if dict = ["able"], then "a4" is not a valid abbreviation since "able" can be abbreviated to "a4".

Your task is to write a function that takes the target word and the dictionary as input and returns the shortest unique abbreviation of the target word. If no such abbreviation exists, return the original word.

Constraints:

The length of the target word and the words in the dictionary will be between 1 and 10.
The dictionary will contain unique words.

Example:

target = "apple", dict = ["blade", "plain", "amber"]
Output: "1p3"

target = "apple", dict = ["able", "plink", "amber"]
Output: "ap3"

How would you approach this problem, and what is the time and space complexity of your solution?

Minimum Unique Word Abbreviation

Hard

Google

1 view

Topics:

StringsBit Manipulation

You are given a target word and a dictionary of words. Your goal is to find the shortest unique abbreviation of the target word such that it does not conflict with any other word in the dictionary.

Abbreviation: An abbreviation can be formed by replacing any non-overlapping substring with the count of characters in that substring. For example, "apple" can be abbreviated as:
- "apple" (no abbreviation)
- "a4" (replace "pple" with 4)
- "4e" (replace "appl" with 4)
- "ap3" (replace "ple" with 3)
- "5" (replace "apple" with 5)
- "a3e" (replace "ppl" with 3)
Conflict: An abbreviation conflicts with a word in the dictionary if the word can be abbreviated to the same string.

Constraints:

The length of the target word and the words in the dictionary will be between 1 and 10.
The dictionary will contain unique words.

Example:

target = "apple", dict = ["blade", "plain", "amber"]
Output: "1p3"

target = "apple", dict = ["able", "plink", "amber"]
Output: "ap3"

How would you approach this problem, and what is the time and space complexity of your solution?

Solution

Minimum Unique Word Abbreviation

Let's explore how to solve the "Minimum Unique Word Abbreviation" problem. The goal is to find the shortest abbreviation of a given word that doesn't conflict with any other word in a dictionary.

1. Naive (Brute Force) Approach

A straightforward but inefficient approach is to generate all possible abbreviations of the target word, then check if each abbreviation is unique (i.e., not an abbreviation of any other word in the dictionary). We can start with shorter abbreviations and increase their length until we find a unique one.

How it Works:
1. Generate all possible abbreviations of length 1, 2, 3, ..., up to the full length of the word.
2. For each abbreviation, check if it's an abbreviation of any word in the dictionary.
3. If an abbreviation is unique, return it.
Example:

Let's say target = "apple" and dict = ["blade"]
1. Abbreviations of length 1: "a", "p", "p", "l", "e"
2. Abbreviations of length 2: "ap", "pl", "le", "a1", "1e", etc.
3. And so on...
Problems: This approach is computationally expensive because generating all possible abbreviations can be exponential in the length of the word. Also, checking if one word is an abbreviation of another involves string comparisons, further increasing the time complexity.

2. Optimal Solution (Bit Manipulation and Pruning)

A more efficient approach involves bit manipulation and pruning to reduce the search space. The main idea is to represent each abbreviation as a bitmask, where each bit represents whether a character in the original word is kept or replaced by a count.

How it Works:
1. Represent Words as Bitmasks: Convert each word in the dictionary into a bitmask. If a word in the dictionary can be abbreviated to the same length as the target, and if they differ at certain positions, we can mark these differences as '1' bits in a mask. Essentially, this tells us the locations where the target word must keep its characters to distinguish itself from other words.
2. Generate Abbreviation Bitmasks: Generate all possible bitmasks for the target word. Each bitmask represents a possible abbreviation.
3. Check Uniqueness: For each generated bitmask, check if it is a valid abbreviation and if it is unique (i.e., doesn't conflict with any word in the dictionary based on the bitmasks).
4. Minimize Length: Choose the shortest unique abbreviation. We can use a Breadth-First Search (BFS) approach to explore the abbreviations in increasing order of the number of kept characters (bits set to 1 in the mask), which helps to find the minimum length abbreviation early.
Example:

`target =

Solution

Minimum Unique Word Abbreviation

Let's explore how to solve the "Minimum Unique Word Abbreviation" problem. The goal is to find the shortest abbreviation of a given word that doesn't conflict with any other word in a dictionary.

1. Naive (Brute Force) Approach

How it Works:
1. Generate all possible abbreviations of length 1, 2, 3, ..., up to the full length of the word.
2. For each abbreviation, check if it's an abbreviation of any word in the dictionary.
3. If an abbreviation is unique, return it.
Example:

Let's say target = "apple" and dict = ["blade"]
1. Abbreviations of length 1: "a", "p", "p", "l", "e"
2. Abbreviations of length 2: "ap", "pl", "le", "a1", "1e", etc.
3. And so on...
Problems: This approach is computationally expensive because generating all possible abbreviations can be exponential in the length of the word. Also, checking if one word is an abbreviation of another involves string comparisons, further increasing the time complexity.

2. Optimal Solution (Bit Manipulation and Pruning)

How it Works:
1. Represent Words as Bitmasks: Convert each word in the dictionary into a bitmask. If a word in the dictionary can be abbreviated to the same length as the target, and if they differ at certain positions, we can mark these differences as '1' bits in a mask. Essentially, this tells us the locations where the target word must keep its characters to distinguish itself from other words.
2. Generate Abbreviation Bitmasks: Generate all possible bitmasks for the target word. Each bitmask represents a possible abbreviation.
3. Check Uniqueness: For each generated bitmask, check if it is a valid abbreviation and if it is unique (i.e., doesn't conflict with any word in the dictionary based on the bitmasks).
4. Minimize Length: Choose the shortest unique abbreviation. We can use a Breadth-First Search (BFS) approach to explore the abbreviations in increasing order of the number of kept characters (bits set to 1 in the mask), which helps to find the minimum length abbreviation early.
Example:

`target =