320. Generalized Abbreviation 🔒

Description

A word's generalized abbreviation can be constructed by taking any number of non-overlapping and non-adjacent substrings and replacing them with their respective lengths.

For example, "abcde" can be abbreviated into:
- "a3e" ("bcd" turned into "3")
- "1bcd1" ("a" and "e" both turned into "1")
- "5" ("abcde" turned into "5")
- "abcde" (no substrings replaced)
However, these abbreviations are invalid:
- "23" ("ab" turned into "2" and "cde" turned into "3") is invalid as the substrings chosen are adjacent.
- "22de" ("ab" turned into "2" and "bc" turned into "2") is invalid as the substring chosen overlap.

Given a string word, return a list of all the possible generalized abbreviations of word. Return the answer in any order.

Example 1:

Input: word = "word"
Output: ["4","3d","2r1","2rd","1o2","1o1d","1or1","1ord","w3","w2d","w1r1","w1rd","wo2","wo1d","wor1","word"]

Example 2:

Input: word = "a"
Output: ["1","a"]

Constraints:

1 <= word.length <= 15
word consists of only lowercase English letters.

Solutions

Solution 1: DFS

We design a function $dfs(i)$, which returns all possible abbreviations for the string $word[i:]$.

The execution logic of the function $dfs(i)$ is as follows:

If $i \geq n$, it means that the string $word$ has been processed, and we directly return a list composed of an empty string.

Otherwise, we can choose to keep $word[i]$, and then add $word[i]$ to the front of each string in the list returned by $dfs(i + 1)$, and add the obtained result to the answer.

We can also choose to delete $word[i]$ and some characters after it. Suppose we delete $word[i..j)$, then the $j$ th character is not deleted, and then add $j - i$ to the front of each string in the list returned by $dfs(j + 1)$, and add the obtained result to the answer.

Finally, we call $dfs(0)$ in the main function.

The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $word$.

Python3JavaC++GoTypeScript

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

        n = len(word)
        return dfs(0)

class Solution {
    private String word;
    private int n;

    public List<String> generateAbbreviations(String word) {
        this.word = word;
        n = word.length();
        return dfs(0);
    }

    private List<String> dfs(int i) {
        if (i >= n) {
            return List.of("");
        }
        List<String> ans = new ArrayList<>();
        for (String s : dfs(i + 1)) {
            ans.add(String.valueOf(word.charAt(i)) + s);
        }
        for (int j = i + 1; j <= n; ++j) {
            for (String s : dfs(j + 1)) {
                ans.add((j - i) + "" + (j < n ? String.valueOf(word.charAt(j)) : "") + s);
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<string> generateAbbreviations(string word) {
        int n = word.size();
        function<vector<string>(int)> dfs = [&](int i) -> vector<string> {
            if (i >= n) {
                return {""};
            }
            vector<string> ans;
            for (auto& s : dfs(i + 1)) {
                string p(1, word[i]);
                ans.emplace_back(p + s);
            }
            for (int j = i + 1; j <= n; ++j) {
                for (auto& s : dfs(j + 1)) {
                    string p = j < n ? string(1, word[j]) : "";
                    ans.emplace_back(to_string(j - i) + p + s);
                }
            }
            return ans;
        };
        return dfs(0);
    }
};

func generateAbbreviations(word string) []string {
    n := len(word)
    var dfs func(int) []string
    dfs = func(i int) []string {
        if i >= n {
            return []string{""}
        }
        ans := []string{}
        for _, s := range dfs(i + 1) {
            ans = append(ans, word[i:i+1]+s)
        }
        for j := i + 1; j <= n; j++ {
            for _, s := range dfs(j + 1) {
                p := ""
                if j < n {
                    p = word[j : j+1]
                }
                ans = append(ans, strconv.Itoa(j-i)+p+s)
            }
        }
        return ans
    }
    return dfs(0)
}

function generateAbbreviations(word: string): string[] {
    const n = word.length;
    const dfs = (i: number): string[] => {
        if (i >= n) {
            return [''];
        }
        const ans: string[] = [];
        for (const s of dfs(i + 1)) {
            ans.push(word[i] + s);
        }
        for (let j = i + 1; j <= n; ++j) {
            for (const s of dfs(j + 1)) {
                ans.push((j - i).toString() + (j < n ? word[j] : '') + s);
            }
        }
        return ans;
    };
    return dfs(0);
}

Solution 2: Binary Enumeration

Since the length of the string $word$ does not exceed $15$, we can use the method of binary enumeration to enumerate all abbreviations. We use a binary number $i$ of length $n$ to represent an abbreviation, where $0$ represents keeping the corresponding character, and $1$ represents deleting the corresponding character. We enumerate all $i$ in the range of $[0, 2^n)$, convert it into the corresponding abbreviation, and add it to the answer list.