Minimum Unique Word Abbreviation

This problem asks to find the shortest abbreviation of a target string that is not an abbreviation of any string in a given dictionary. The challenge lies in efficiently generating and checking abbreviations, considering the constraints on abbreviation length and the potentially large dictionary.

Approach:

The solution employs a backtracking approach to generate all possible abbreviations of the target string. For each abbreviation, it checks if it's unique (i.e., not an abbreviation of any word in the dictionary). The algorithm keeps track of the shortest unique abbreviation found so far.

Algorithm:

1. Abbreviation Generation: The core logic recursively generates abbreviations. It iterates through the target string, choosing at each position whether to include the character or to replace a substring with its length. This is done using bit manipulation. Each bit in a mask represents a choice: 0 means include the character, 1 means replace a substring. Adjacent 1s are invalid because they violate the non-adjacent substrings constraint.

2. Uniqueness Check: For each generated abbreviation, the algorithm checks if it's an abbreviation of any word in the dictionary. This is done by comparing the abbreviation against each dictionary word.

3. Shortest Abbreviation: The algorithm keeps track of the shortest unique abbreviation encountered. If a shorter unique abbreviation is found, it updates the result.

Time Complexity Analysis:

• Abbreviation Generation: The number of possible abbreviations is exponential with respect to the length of the target string (m). In the worst case, you'd explore all possible combinations of including/replacing characters. This gives us a time complexity of O(2m) where m is the length of the target string.

• Uniqueness Check: Checking if an abbreviation is unique involves comparing it against each word in the dictionary (n words). This adds a factor of O(n) to the overall time complexity.

• Total Time Complexity: Combining the generation and uniqueness check, the total time complexity is O(n * 2m). This makes the algorithm exponential and may be slow for very long target strings.

Space Complexity Analysis:

The space complexity is primarily determined by the recursive call stack in the backtracking function, which can be proportional to the length of the target string (m) in the worst case. Additionally, storing the shortest unique abbreviation and other intermediate results might add a small constant amount of space. Therefore, the space complexity is O(m).

Code (Python):

def minUniqueWordAbbreviation(target, dictionary):
    """
    Finds the shortest unique abbreviation of the target string.
 
    Args:
        target: The target string.
        dictionary: A list of strings.
 
    Returns:
        The shortest unique abbreviation, or None if none found.
    """
 
    shortest_unique_abbr = ""
 
    def is_abbreviation(word, abbr):
        """Checks if word is an abbreviation of abbr."""
        i = 0
        j = 0
        while i < len(word) and j < len(abbr):
            if abbr[j].isalpha():
                if abbr[j] != word[i]:
                    return False
                i += 1
                j += 1
            else:
                num = 0
                while j < len(abbr) and abbr[j].isdigit():
                    num = num * 10 + int(abbr[j])
                    j += 1
                i += num
        return i == len(word) and j == len(abbr)
 
 
    def generate_abbreviation(index, current_abbr, mask):
        nonlocal shortest_unique_abbr
        if index == len(target):
            unique = True
            for word in dictionary:
                if is_abbreviation(word, current_abbr):
                    unique = False
                    break
            if unique:
                if not shortest_unique_abbr or len(current_abbr) < len(shortest_unique_abbr):
                    shortest_unique_abbr = current_abbr
            return
 
        generate_abbreviation(index + 1, current_abbr + target[index], mask << 1)  # Include character
 
        count = 0
        next_index = index + 1
        while next_index < len(target) and (mask & 1) == 0:  # Check for adjacent 1s
            count += 1
            next_index +=1
            mask <<=1
        if next_index <= len(target):
            generate_abbreviation(next_index, current_abbr + str(count + 1), mask << 1)  # Replace substring
 
    generate_abbreviation(0, "", 0)
    return shortest_unique_abbr
 
#Example
target = "apple"
dictionary = ["blade", "plain", "amber"]
result = minUniqueWordAbbreviation(target, dictionary)
print(f"Shortest unique abbreviation for '{target}': {result}")
 

Note: The Python code provides a functional solution. Optimizations could be explored to further improve performance for larger inputs (e.g., more sophisticated pruning techniques in the backtracking). Other languages (Java, C++, Go) would follow a similar backtracking and abbreviation-checking structure. The key would be to implement the is_abbreviation function effectively to handle the substring replacement logic efficiently.