LeetCode 320: Generalized Abbreviation
LeetCode 320 Solution 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.
: :
LeetCode 320 Solutions in Java, C++, Python
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);
}
}
}Interactive Code Editor for LeetCode 320
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