Solution Explanation: Generalized Abbreviation

The problem asks to generate all possible generalized abbreviations of a given string. A generalized abbreviation replaces one or more consecutive characters with their count. For example, "word" can be abbreviated as "word", "4", "w3", "wo2", etc. The key constraint is that the replaced substrings cannot overlap or be adjacent.

Two main approaches are presented to solve this: Depth-First Search (DFS) and Binary Enumeration.

Approach 1: Depth-First Search (DFS)

This recursive approach explores all possible abbreviation combinations. The dfs(i) function generates abbreviations for the substring starting from index i.

Logic:

1. Base Case: If i reaches the end of the string, return a list containing an empty string (representing the end of an abbreviation).

2. Recursive Steps: For each character at index i:

   • Keep: Add the character at i to the beginning of each abbreviation generated by recursively calling dfs(i+1) and add to the results.
   • Skip: Iterate from i+1 to the end of the string. For each j, consider skipping characters from i to j-1 and replacing them with their count (j-i). Recursively call dfs(j+1) to generate the remaining abbreviation. Add j-i and, if applicable, the character at index j, to the beginning of each of the returned abbreviation strings.

Time Complexity: O(n * 2n) - In the worst case, for each character, we have two choices (keep or skip), leading to an exponential number of combinations. The work done for each combination is linear in the length of the string (n).

Space Complexity: O(n) - The recursive depth is at most n (length of the string). The space used for the result list can also grow proportionally to the number of possible abbreviations.

Approach 2: Binary Enumeration

This approach utilizes bit manipulation to represent abbreviation choices efficiently. Each bit in a binary number of length n (the string length) corresponds to a character in the input string.

Logic:

1. Iterate through all possible binary numbers from 0 to 2n - 1.
2. Each bit represents a choice: 0 means keep the character, 1 means skip it.
3. For each binary number:
   • Initialize a count to 0.
   • Iterate through the bits:
     • If a bit is 1 (skip), increment the count.
     • If a bit is 0 (keep), add the count to the abbreviation string (if count > 0), reset the count, and add the character.
   • After processing all bits, add the final count (if it is greater than 0) to the abbreviation string.

Time Complexity: O(n * 2n) - We iterate through 2n binary numbers, and for each, we process the string of length n.

Space Complexity: O(n) - The space used by the result list depends on the number of abbreviations, which is at most 2n.

Code Examples (Python3):

Approach 1 (DFS):

def generateAbbreviations(word: str) -> List[str]:
    def dfs(i: int) -> List[str]:
        if i == len(word):
            return [""]
        ans = [word[i] + s for s in dfs(i + 1)]
        for j in range(i + 1, len(word) + 1):
            for s in dfs(j + 1):
                ans.append(str(j - i) + (word[j] if j < len(word) else "") + s)
        return ans
    return dfs(0)
 

Approach 2 (Binary Enumeration):

def generateAbbreviations(word: str) -> List[str]:
    n = len(word)
    ans = []
    for i in range(1 << n):
        cnt = 0
        s = []
        for j in range(n):
            if i >> j & 1:
                cnt += 1
            else:
                if cnt:
                    s.append(str(cnt))
                    cnt = 0
                s.append(word[j])
        if cnt:
            s.append(str(cnt))
        ans.append("".join(s))
    return ans
 

Both approaches achieve the same result, but binary enumeration might be slightly more concise. The choice depends on preference and familiarity with bit manipulation. The time complexity is inherently exponential due to the nature of the problem.