Generalized Abbreviation

class Solution {
public:
    vector<string> abbreviateWord(string text) {
        int len = text.length();
        vector<string> result;

        for (int code = 0; code < (1 << len); code++) {
            string temp;
            int count = 0;

            for (int i = 0; i < len; i++) {
                if (code & (1 << i)) {
                    count++;
                } else {
                    if (count != 0) {
                        temp += to_string(count);
                    }
                    temp += text[i];
                    count = 0;
                }
            }

            if (count > 0) {
                temp += to_string(count);
            }
            result.push_back(temp);
        }

        return result;
    }
};

This C++ code defines a solution to generate all possible abbreviations of a given word. The process involves using bit manipulation to explore each potential abbreviation combination:

• The abbreviateWord function takes a string text as input and initializes a vector result to store all abbreviations.
• It calculates the length of the input string and iterates through all possible combinations using a bit masking approach (1 << len gives the number of combinations).
• In each iteration, the code builds a temporary abbreviation string temp. A counter count tracks the number of consecutive characters that are abbreviated.
• For every bit in the combination (code), it checks if the current bit is set, indicating that the corresponding character in the text should be abbreviated. If so, it increments count by one.
• If the bit is not set (character is not abbreviated), it first adds the count of abbreviated characters (if any) as a numeral in temp, then adds the current character from text and resets count.
• After processing all bits (characters in the word), if any characters have been abbreviated at the end (count > 0), it adds the number to temp.
• The constructed abbreviation temp is then added to result.
• Finally, the function returns the vector result containing all possible abbreviations for the input text.

This implementation efficiently generates all possible generalized abbreviations of a given word using bitwise operations to handle each character's abbreviation status.

class Solution {

    public List<String> getAbbrList(String input) {
        int len = input.length();
        List<String> result = new ArrayList<>();

        for (int m = 0; m < (1 << len); m++) {
            StringBuilder abbr = new StringBuilder();
            int count = 0;

            for (int i = 0; i < len; i++) {
                if ((m & (1 << i)) != 0) {
                    count++;
                } else {
                    if (count > 0) {
                        abbr.append(count);
                        count = 0;
                    }
                    abbr.append(input.charAt(i));
                }
            }

            if (count > 0) {
                abbr.append(count);
            }
            result.add(abbr.toString());
        }

        return result;
    }
}

The solution provided generates all possible generalized abbreviations for a given string using a bit manipulation approach in Java. Begin by passing a string to the getAbbrList method. This method calculates the length of the string and initializes a list to store the result.

Employ a loop that iterates over all possible combinations of characters and abbreviations for the string. This is achieved using a binary representation where each bit determines whether to abbreviate (as a count of consecutive characters) or to keep the character as is.

Inside the loop, initialize a counter for counting consecutive characters to abbreviate and a StringBuilder for constructing the current abbreviation. Another nested loop iterates over each character in the string, checking the respective bit in the current bitmask. If the bit is set, increment the counter; otherwise, append the accumulated count as a number (if any) followed by the character itself, then reset the counter.

After exiting the inner loop, check if there remains any count of characters to abbreviate and, if so, append this count to the abbreviation. Finally, add the constructed abbreviation string to the result list.

Return the list after processing all bit combinations. Each string in the returned list represents a unique way of abbreviating the input string.

This solution efficiently explores all potential abbreviations using bitwise operations to manage combinations, applicable to scenarios where such transformations are needed, e.g., in compression algorithms or generating unique identifiers.

class Solution:
    def getAbbreviations(self, txt):
        length = len(txt)
        result = []

        for i in range(1 << length):
            temp = []
            count = 0

            for j in range(length):
                if i & (1 << j):
                    count += 1
                else:
                    if count:
                        temp.append(str(count))
                        count = 0
                    temp.append(txt[j])

            if count:
                temp.append(str(count))

            result.append("".join(temp))

        return result

The provided Python code solves the problem of generating all possible generalized abbreviations of a given string. It employs a bit manipulation technique to explore all combinations of abbreviated and non-abbreviated characters. Here's a breakdown of how the solution is implemented:

• The main function defined in the Solution class is getAbbreviations, which takes txt as input.
• First, it captures the length of txt and initializes an empty list result to hold all possible abbreviations.
• It uses a loop to iterate through all possible binary numbers from 0 to 2^length - 1. Each binary number represents a unique combination of abbreviating or not abbreviating each character in txt.
• Inside this loop, another loop iterates through each bit of the binary number:
  • If the bit is set (1), it increases a counter count that tracks the number of consecutive characters to abbreviate.
  • If the bit is clear (0), it checks if count is nonzero. If so, it appends the count to temp as a string, resets count, and appends the current character from txt.
• After processing all characters, if count is still nonzero, it appends this count to temp as the final abbreviation count for any trailing characters.
• Each constructed abbreviation from temp (joined into a string) is added to the result list.
• Finally, the function returns the result list containing all possible abbreviations.

This code efficiently generates abbreviations by cleverly using binary numbers to represent whether each character in the string should be abbreviated or written as is. This method dramatically simplifies the complexity of handling nested conditions and manually tracking segments of the string to abbreviate.