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 $word$.

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.

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

• class Solution {
public List<String> generateAbbreviations(String word) {
int n = word.length();
List<String> ans = new ArrayList<>();
for (int i = 0; i < 1 << n; ++i) {
StringBuilder s = new StringBuilder();
int cnt = 0;
for (int j = 0; j < n; ++j) {
if ((i >> j & 1) == 1) {
++cnt;
} else {
if (cnt > 0) {
s.append(cnt);
cnt = 0;
}
s.append(word.charAt(j));
}
}
if (cnt > 0) {
s.append(cnt);
}
}
return ans;
}
}

• class Solution {
public:
vector<string> generateAbbreviations(string word) {
int n = word.size();
vector<string> ans;
for (int i = 0; i < 1 << n; ++i) {
string s;
int cnt = 0;
for (int j = 0; j < n; ++j) {
if (i >> j & 1) {
++cnt;
} else {
if (cnt) {
s += to_string(cnt);
cnt = 0;
}
s.push_back(word[j]);
}
}
if (cnt) {
s += to_string(cnt);
}
ans.push_back(s);
}
return ans;
}
};

• class Solution:
def generateAbbreviations(self, word: str) -> List[str]:
n = len(word)
ans = []
for i in range(1 << n):
cnt = 0
s = []
for j in range(n):
if i >> j & 1:
cnt += 1
else:
if cnt:
s.append(str(cnt))
cnt = 0
s.append(word[j])
if cnt:
s.append(str(cnt))
ans.append("".join(s))
return ans


• func generateAbbreviations(word string) (ans []string) {
n := len(word)
for i := 0; i < 1<<n; i++ {
s := &strings.Builder{}
cnt := 0
for j := 0; j < n; j++ {
if i>>j&1 == 1 {
cnt++
} else {
if cnt > 0 {
s.WriteString(strconv.Itoa(cnt))
cnt = 0
}
s.WriteByte(word[j])
}
}
if cnt > 0 {
s.WriteString(strconv.Itoa(cnt))
}
ans = append(ans, s.String())
}
return
}

• function generateAbbreviations(word: string): string[] {
const n = word.length;
const ans: string[] = [];
for (let i = 0; i < 1 << n; ++i) {
const s: string[] = [];
let cnt = 0;
for (let j = 0; j < n; ++j) {
if ((i >> j) & 1) {
++cnt;
} else {
if (cnt > 0) {
s.push(cnt.toString());
cnt = 0;
}
s.push(word[j]);
}
}
if (cnt > 0) {
s.push(cnt.toString());
}
ans.push(s.join(''));
}
return ans;
}