LeetCode 320: Generalized Abbreviation Solution

Master LeetCode problem 320 (Generalized Abbreviation), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

320. Generalized Abbreviation

Medium

String Backtracking Bit Manipulation

Problem Explanation

Explanation:

This problem asks us to generate all possible abbreviations of a given word. An abbreviation can be formed by replacing one or more characters in the word with a digit representing the number of skipped characters. The goal is to generate all possible abbreviations of the word.

We can solve this problem using backtracking. We will explore two choices at each step: either abbreviate the next character or keep it as it is. We will recursively generate all possible abbreviations and backtrack to explore other choices.

Algorithm:

Start with an empty string and an index pointing to the beginning of the word.
At each step, we have two choices:
- Abbreviate the current character by increasing the count of skipped characters.
- Keep the current character without abbreviation.
Recursively explore both choices for the next character.
If we reach the end of the word, add the generated abbreviation to the result.
Backtrack to explore other choices by removing the last abbreviation and continue the exploration.

Time Complexity:

The time complexity of this approach is O(2^N) where N is the length of the input word. This is because for each character, we have two choices - either abbreviate it or keep it as it is.

Space Complexity:

The space complexity is also O(2^N) considering the recursive stack space used during backtracking.

: :

Solution Code

class Solution {
    public List<String> generateAbbreviations(String word) {
        List<String> result = new ArrayList<>();
        backtrack(result, word, 0, "", 0);
        return result;
    }
    
    private void backtrack(List<String> result, String word, int index, String current, int count) {
        if (index == word.length()) {
            if (count > 0) {
                current += count;
            }
            result.add(current);
        } else {
            backtrack(result, word, index + 1, current + (count > 0 ? count : "") + word.charAt(index), 0);
            backtrack(result, word, index + 1, current, count + 1);
        }
    }
}

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