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Generalized Abbreviation

Medium

Asked by:

12 views

Topics:

StringsRecursion

Write a function to generate all possible abbreviations of a word.

Note: The order of the output does not matter.

Example:

Input: "word"
Output:
[
  "word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"
]

Try coding it in LeetCode

Generalized Abbreviation

Medium

Asked by:

12 views

Topics:

StringsRecursion

Write a function to generate all possible abbreviations of a word.

Note: The order of the output does not matter.

Example:

Input: "word"
Output:
[
  "word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"
]

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input string contain characters other than lowercase English letters?
What should I return if the input string is null or empty?
Are there any constraints on the length of the input string?
Is the order of the generated abbreviations important in the output?
Should the output contain duplicate abbreviations, or should it be a set of unique abbreviations?

Brute Force Solution

Approach

The brute force approach for generalized abbreviation means trying every possible combination of abbreviating the input word. We explore each choice: either abbreviate a character or keep it as is. This generates all possible abbreviations, which we then save as solutions.

Here's how the algorithm would work step-by-step:

Start with the original word as the first possible abbreviation.
For each letter in the word, consider two options: keep the letter as it is or abbreviate it by representing it as a number.
If you keep the letter, move on to the next letter and repeat the process.
If you abbreviate the letter, represent it as '1' (meaning one letter is abbreviated), and move to the next letter.
If consecutive letters are abbreviated, merge the consecutive '1's into a single number that represents the total number of consecutive abbreviated letters.
Keep track of all these possible abbreviated forms.
Once you've gone through all the letters, you have a complete abbreviation. Save this abbreviation.
Repeat this process until you've explored every possible combination of abbreviating or not abbreviating each letter.
The collection of all saved abbreviations is your final result.

Code Implementation

def generalized_abbreviation_brute_force(word):
    result = []
    
    def backtrack(index, current_abbreviation, consecutive_abbreviation_count):
        # Base case: we've processed all characters in the word
        if index == len(word):
            if consecutive_abbreviation_count > 0:
                current_abbreviation += str(consecutive_abbreviation_count)
            result.append(current_abbreviation)
            return

        # Option 1: Abbreviate the current character
        # Explore abbreviating the current character.
        backtrack(index + 1, current_abbreviation,
                  consecutive_abbreviation_count + 1)

        # Option 2: Keep the current character
        # If we had consecutive abbreviations, add count.
        if consecutive_abbreviation_count > 0:
            backtrack(index + 1, current_abbreviation + \
                      str(consecutive_abbreviation_count) + word[index], 0)

        # No consecutive abbreviations, add letter
        else:
            backtrack(index + 1, current_abbreviation + word[index], 0)

    backtrack(0, "", 0)
    return result

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores every possible combination of abbreviating or not abbreviating each letter in the input word of length n. For each letter, there are two choices: either abbreviate it or keep it as is. This branching at each letter leads to 2 * 2 * ... * 2 (n times) possibilities, which equals 2^n. Therefore, the time complexity is O(2^n) because the algorithm essentially generates all possible subsets of the input string’s characters corresponding to whether to abbreviate or not.

Space Complexity

O(N) – The provided solution explores all possible abbreviations of the input word using a recursive approach. The depth of the recursion can go up to N, where N is the length of the word, representing the number of characters. Each recursive call adds a frame to the call stack, thus the recursion stack can grow to a maximum depth of N. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The optimal way to generate generalized abbreviations is to systematically explore all possible combinations of abbreviating or not abbreviating each character in the input. We achieve this through a recursive approach, making a choice at each character to either abbreviate it (incrementing a count) or keep it as is (adding the character to our result).

Here's how the algorithm would work step-by-step:

Start with an empty abbreviation and a counter for consecutive abbreviated characters, beginning at the first character of the word.
At each character, you have two choices: abbreviate it or keep it.
If you choose to abbreviate, increment the abbreviation counter.
If you choose to keep, add the current count (if it's greater than 0) and the current character to the abbreviation.
Move to the next character in the word and repeat the process.
When you reach the end of the word, add the final count (if greater than 0) to the abbreviation, and that's one possible abbreviation.
Repeat this process, exploring all possible choices at each character, to generate all the abbreviations.

Code Implementation

def generate_abbreviations(word):
    result = []

    def backtrack(index, current_abbreviation, abbreviation_count):
        if index == len(word):
            # Append any remaining count to the abbreviation.
            if abbreviation_count > 0:
                current_abbreviation += str(abbreviation_count)
            result.append(current_abbreviation)
            return

        # Option 1: Abbreviate the current character.
        backtrack(index + 1, current_abbreviation, abbreviation_count + 1)

        # Option 2: Keep the current character.
        # Append the count before the char.
        next_abbreviation = current_abbreviation
        if abbreviation_count > 0:
            next_abbreviation += str(abbreviation_count)

        backtrack(index + 1, next_abbreviation + word[index], 0)

    backtrack(0, "", 0)
    return result

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible combinations of abbreviating or not abbreviating each character in the input string of length n. For each character, there are two choices: abbreviate or not. This results in a binary tree of decisions with a depth of n. Therefore, the total number of possible abbreviations (and thus the number of nodes visited in the implicit decision tree) is 2^n. The complexity is thus O(2^n).

Space Complexity

O(N) – The space complexity is dominated by the recursion depth, where N is the length of the input word. In the worst-case scenario, where no characters are abbreviated, each recursive call adds a new frame to the call stack. Thus, the maximum depth of the recursion is N, leading to a space complexity of O(N) due to the call stack. Although the string building within each call technically adds to the space, the depth of the recursion dominates the total auxiliary space used.

Edge Cases

Case	How to Handle
Null or empty input string	Return a list containing an empty string, as that's the only possible abbreviation of an empty string.
String containing non-alphanumeric characters	Treat all characters equally and abbreviate them numerically if needed without special handling.
String with only one character	Return a list containing the original string and '1'.
Very long input string	The solution's space and time complexity should scale reasonably (e.g., O(2^n) where n is the string length), consider using generators for large n.
String with repeating characters like 'aaaa'	Ensure that abbreviation logic correctly handles consecutive repeated characters (e.g., aaaa can be '4' or 'a3', '2a2' etc.).
String with numbers as characters	Treat numeric characters like any other character; ensure correct abbreviation calculation (e.g., 'a1a' becomes 'a1a', 'a2', '11a', '3').
Recursion depth exceeding system limits	If using recursion, optimize tail recursion, or convert to iterative approach to prevent stack overflow for very long strings.
Memory constraints with large output list	Consider a generator-based approach to yield abbreviations one at a time instead of storing all in memory at once.

Null or empty input string

How to Handle:

Return a list containing an empty string, as that's the only possible abbreviation of an empty string.

String containing non-alphanumeric characters

How to Handle:

Treat all characters equally and abbreviate them numerically if needed without special handling.

String with only one character

How to Handle:

Return a list containing the original string and '1'.

Very long input string

How to Handle:

The solution's space and time complexity should scale reasonably (e.g., O(2^n) where n is the string length), consider using generators for large n.

String with repeating characters like 'aaaa'

How to Handle:

Ensure that abbreviation logic correctly handles consecutive repeated characters (e.g., aaaa can be '4' or 'a3', '2a2' etc.).

String with numbers as characters

How to Handle:

Treat numeric characters like any other character; ensure correct abbreviation calculation (e.g., 'a1a' becomes 'a1a', 'a2', '11a', '3').

Recursion depth exceeding system limits

How to Handle:

If using recursion, optimize tail recursion, or convert to iterative approach to prevent stack overflow for very long strings.

Memory constraints with large output list

How to Handle:

Consider a generator-based approach to yield abbreviations one at a time instead of storing all in memory at once.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input string contain characters other than lowercase English letters?
What should I return if the input string is null or empty?
Are there any constraints on the length of the input string?
Is the order of the generated abbreviations important in the output?
Should the output contain duplicate abbreviations, or should it be a set of unique abbreviations?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the original word as the first possible abbreviation.
For each letter in the word, consider two options: keep the letter as it is or abbreviate it by representing it as a number.
If you keep the letter, move on to the next letter and repeat the process.
If you abbreviate the letter, represent it as '1' (meaning one letter is abbreviated), and move to the next letter.
If consecutive letters are abbreviated, merge the consecutive '1's into a single number that represents the total number of consecutive abbreviated letters.
Keep track of all these possible abbreviated forms.
Once you've gone through all the letters, you have a complete abbreviation. Save this abbreviation.
Repeat this process until you've explored every possible combination of abbreviating or not abbreviating each letter.
The collection of all saved abbreviations is your final result.

Code Implementation

def generalized_abbreviation_brute_force(word):
    result = []
    
    def backtrack(index, current_abbreviation, consecutive_abbreviation_count):
        # Base case: we've processed all characters in the word
        if index == len(word):
            if consecutive_abbreviation_count > 0:
                current_abbreviation += str(consecutive_abbreviation_count)
            result.append(current_abbreviation)
            return

        # Option 1: Abbreviate the current character
        # Explore abbreviating the current character.
        backtrack(index + 1, current_abbreviation,
                  consecutive_abbreviation_count + 1)

        # Option 2: Keep the current character
        # If we had consecutive abbreviations, add count.
        if consecutive_abbreviation_count > 0:
            backtrack(index + 1, current_abbreviation + \
                      str(consecutive_abbreviation_count) + word[index], 0)

        # No consecutive abbreviations, add letter
        else:
            backtrack(index + 1, current_abbreviation + word[index], 0)

    backtrack(0, "", 0)
    return result

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start with an empty abbreviation and a counter for consecutive abbreviated characters, beginning at the first character of the word.
At each character, you have two choices: abbreviate it or keep it.
If you choose to abbreviate, increment the abbreviation counter.
If you choose to keep, add the current count (if it's greater than 0) and the current character to the abbreviation.
Move to the next character in the word and repeat the process.
When you reach the end of the word, add the final count (if greater than 0) to the abbreviation, and that's one possible abbreviation.
Repeat this process, exploring all possible choices at each character, to generate all the abbreviations.

Code Implementation

def generate_abbreviations(word):
    result = []

    def backtrack(index, current_abbreviation, abbreviation_count):
        if index == len(word):
            # Append any remaining count to the abbreviation.
            if abbreviation_count > 0:
                current_abbreviation += str(abbreviation_count)
            result.append(current_abbreviation)
            return

        # Option 1: Abbreviate the current character.
        backtrack(index + 1, current_abbreviation, abbreviation_count + 1)

        # Option 2: Keep the current character.
        # Append the count before the char.
        next_abbreviation = current_abbreviation
        if abbreviation_count > 0:
            next_abbreviation += str(abbreviation_count)

        backtrack(index + 1, next_abbreviation + word[index], 0)

    backtrack(0, "", 0)
    return result

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return a list containing an empty string, as that's the only possible abbreviation of an empty string.
String containing non-alphanumeric characters	Treat all characters equally and abbreviate them numerically if needed without special handling.
String with only one character	Return a list containing the original string and '1'.
Very long input string	The solution's space and time complexity should scale reasonably (e.g., O(2^n) where n is the string length), consider using generators for large n.
String with repeating characters like 'aaaa'	Ensure that abbreviation logic correctly handles consecutive repeated characters (e.g., aaaa can be '4' or 'a3', '2a2' etc.).
String with numbers as characters	Treat numeric characters like any other character; ensure correct abbreviation calculation (e.g., 'a1a' becomes 'a1a', 'a2', '11a', '3').
Recursion depth exceeding system limits	If using recursion, optimize tail recursion, or convert to iterative approach to prevent stack overflow for very long strings.
Memory constraints with large output list	Consider a generator-based approach to yield abbreviations one at a time instead of storing all in memory at once.

Null or empty input string

How to Handle:

Return a list containing an empty string, as that's the only possible abbreviation of an empty string.

String containing non-alphanumeric characters

How to Handle:

Treat all characters equally and abbreviate them numerically if needed without special handling.

String with only one character

How to Handle:

Return a list containing the original string and '1'.

Very long input string

How to Handle:

The solution's space and time complexity should scale reasonably (e.g., O(2^n) where n is the string length), consider using generators for large n.

String with repeating characters like 'aaaa'

How to Handle:

Ensure that abbreviation logic correctly handles consecutive repeated characters (e.g., aaaa can be '4' or 'a3', '2a2' etc.).

String with numbers as characters

How to Handle:

Treat numeric characters like any other character; ensure correct abbreviation calculation (e.g., 'a1a' becomes 'a1a', 'a2', '11a', '3').

Recursion depth exceeding system limits

How to Handle:

If using recursion, optimize tail recursion, or convert to iterative approach to prevent stack overflow for very long strings.

Memory constraints with large output list

How to Handle:

Consider a generator-based approach to yield abbreviations one at a time instead of storing all in memory at once.