Solution Approach

The implementation uses a depth-first search (DFS) approach with a recursive helper function dfs(i) that generates all possible abbreviations for the substring word[i:].

Base Case: When i >= n (where n is the length of the word), we've processed the entire string. Return a list containing an empty string [""] as there's nothing left to abbreviate.

Recursive Cases:

For each position i, we explore two types of choices:

1. Keep the current character:

   • We keep word[i] as-is
   • Recursively get all abbreviations for word[i+1:] by calling dfs(i + 1)
   • Prepend word[i] to each returned abbreviation
   • Add these results to our answer list: ans = [word[i] + s for s in dfs(i + 1)]

2. Abbreviate a segment starting at position i:

   • Try different segment lengths by iterating j from i+1 to n+1
   • This represents abbreviating characters from index i to j-1 (exclusive of j)
   • The length of the abbreviated segment is j - i
   • After abbreviating [i, j), we continue from position j+1 (not j) to ensure non-adjacent abbreviations
   • For each segment length:
     • Get all abbreviations for the substring starting at j+1 by calling dfs(j + 1)
     • Build the abbreviation by combining:
       • The number str(j - i) (length of abbreviated segment)
       • The character at position j if it exists (word[j] if j < n else "") - this acts as a separator
       • Each abbreviation returned from dfs(j + 1)
     • Add to answer: ans.append(str(j - i) + (word[j] if j < n else "") + s)

Key Implementation Details:

• The function processes the string from left to right, making decisions at each position
• By jumping to j+1 after abbreviating [i, j), we ensure that the character at position j (if it exists) acts as a separator, preventing adjacent abbreviations
• The recursive structure naturally handles all combinations of keeping and abbreviating different parts of the string
• Starting the recursion with dfs(0) generates all abbreviations for the entire word

Example Walkthrough for "ab":

• dfs(0) can:
  • Keep 'a': recurse with dfs(1), which gives ['b', '1'], resulting in ['ab', 'a1']
  • Abbreviate 'a' (segment [0,1)): continue from dfs(2) with 'b' as separator, giving ['1b']
  • Abbreviate 'ab' (segment [0,2)): continue from dfs(3), giving ['2']
• Final result: ['ab', 'a1', '1b', '2']

Example Walkthrough

Let's walk through the algorithm with the string "abc" to see how it generates all possible abbreviations.

Starting with dfs(0) - Processing position 0:

At position 0, we have character 'a'. We can either:

1. Keep 'a': Call dfs(1) to process the rest "bc"

   • From dfs(1) at 'b', we can:
     • Keep 'b': Call dfs(2) → gives us ['c', '1'] → prepend 'b' → ['bc', 'b1']
     • Abbreviate 'b' (length 1): Skip to dfs(3) with 'c' as separator → gives ['1c']
     • Abbreviate 'bc' (length 2): Skip to dfs(4) → gives ['2']
   • So dfs(1) returns: ['bc', 'b1', '1c', '2']
   • Prepending 'a' to each: ['abc', 'ab1', 'a1c', 'a2']

2. Abbreviate starting from position 0:

   • Abbreviate 'a' only (j=1, length=1): Continue from dfs(2) with 'b' as separator

     • dfs(2) at 'c' gives: ['c', '1']
     • Combine: '1' + 'b' + each result → ['1bc', '1b1']

   • Abbreviate 'ab' (j=2, length=2): Continue from dfs(3) with 'c' as separator

     • dfs(3) is beyond bounds, returns ['']
     • Combine: '2' + 'c' + '' → ['2c']

   • Abbreviate 'abc' (j=3, length=3): Continue from dfs(4)

     • dfs(4) is beyond bounds, returns ['']
     • Combine: '3' + '' + '' → ['3']

Combining all results:

• From keeping 'a': ['abc', 'ab1', 'a1c', 'a2']
• From abbreviating 'a': ['1bc', '1b1']
• From abbreviating 'ab': ['2c']
• From abbreviating 'abc': ['3']

Final answer: ['abc', 'ab1', 'a1c', 'a2', '1bc', '1b1', '2c', '3']

Notice how the algorithm naturally prevents adjacent abbreviations. When we abbreviate 'a' (position 0), we jump to position 2, ensuring 'b' acts as a separator. This prevents invalid forms like "11c" which would require adjacent abbreviations of 'a' and 'b'.

Time and Space Complexity

Time Complexity: O(n × 2^n)

The algorithm generates all possible abbreviations of a word by making decisions at each position: either keep the character or abbreviate a sequence of characters. At each position i, we have two main choices:

• Keep the current character and continue (leading to 2^n possible combinations)
• Abbreviate different lengths starting from position i (from 1 to n-i characters)

For each of the 2^n possible abbreviations, we need O(n) time to construct the string. Therefore, the total time complexity is O(n × 2^n).

Space Complexity: O(n)

The space complexity consists of:

• Recursion stack depth: The maximum depth of recursion is n (when we process each character one by one), requiring O(n) stack space
• Temporary strings during construction: While we create multiple strings in the result, we only need to consider the space for the recursion stack and temporary variables at any given moment
• The output list itself contains 2^n strings, but this is typically not counted in space complexity analysis as it's the required output

Therefore, excluding the output storage, the space complexity is O(n).

Common Pitfalls

1. Incorrect Handling of Adjacent Abbreviations

The Pitfall: A common mistake is to continue from position j instead of j+1 after abbreviating the segment [i, j). This would create invalid adjacent abbreviations.

Incorrect Implementation:

# WRONG: This allows adjacent abbreviations for end_index in range(start_index + 1, word_length + 1):  abbreviation_length = end_index - start_index  # Continuing from end_index allows adjacent abbreviations  for suffix in generate_from_index(end_index): # ❌ Should be end_index + 1  abbreviations.append(str(abbreviation_length) + suffix)

Why it's wrong:

• If we abbreviate [0, 2) in "abcde" to get "2", then continue from index 2, we could immediately abbreviate [2, 4) to get "2" again, resulting in "22e"
• This creates adjacent numbers without any character separator, violating the non-adjacent rule

Correct Solution:

# CORRECT: Skip to j+1 to ensure non-adjacent abbreviations for end_index in range(start_index + 1, word_length + 1):  abbreviation_length = end_index - start_index  # Skip the character at end_index (if it exists) to act as separator  for suffix in generate_from_index(end_index + 1): # ✓ Correctly skip one position  if end_index < word_length:  # Include the separator character  abbreviations.append(str(abbreviation_length) + word[end_index] + suffix)  else:  # No separator needed at the end  abbreviations.append(str(abbreviation_length) + suffix)

2. Forgetting to Handle the Separator Character

The Pitfall: Another mistake is forgetting to include the character at position j when building the abbreviation after abbreviating segment [i, j).

Incorrect Implementation:

# WRONG: Missing the separator character for end_index in range(start_index + 1, word_length + 1):  abbreviation_length = end_index - start_index  for suffix in generate_from_index(end_index + 1):  # Missing word[end_index] when it exists  abbreviations.append(str(abbreviation_length) + suffix) # ❌

Why it's wrong:

• For "abc", if we abbreviate "a" (segment [0,1)), we should get "1bc", not "1c"
• The character at position 1 ('b') must be included as it serves as the separator

Correct Solution: Always check if there's a character at position end_index and include it:

if end_index < word_length:  # Include the character at end_index as separator  abbreviations.append(str(abbreviation_length) + word[end_index] + suffix) else:  # We've abbreviated to the end, no separator needed  abbreviations.append(str(abbreviation_length) + suffix)

3. Inefficient String Concatenation

The Pitfall: While not affecting correctness, repeatedly concatenating strings in Python can be inefficient for longer words due to string immutability.

Less Efficient:

abbreviations.append(word[start_index] + suffix) # Creates new string each time

More Efficient Alternative: For performance-critical scenarios with very long words, consider using a list to build the result:

def generate_from_index(start_index: int, current_path: List[str]) -> None:  if start_index >= word_length:  result.append(''.join(current_path))  return   # Keep current character  current_path.append(word[start_index])  generate_from_index(start_index + 1, current_path)  current_path.pop()   # Abbreviate segments  for end_index in range(start_index + 1, word_length + 1):  current_path.append(str(end_index - start_index))  if end_index < word_length:  current_path.append(word[end_index])  generate_from_index(end_index + 1, current_path)  current_path.pop()  else:  generate_from_index(end_index + 1, current_path)  current_path.pop()

This approach modifies a single list rather than creating new strings at each step, which can be more efficient for very long input strings.