Leetcode #320: Generalized Abbreviation
In this guide, we solve Leetcode #320 Generalized Abbreviation in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Bit Manipulation, String, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: word = "word"
Output: ["4","3d","2r1","2rd","1o2","1o1d","1or1","1ord","w3","w2d","w1r1","w1rd","wo2","wo1d","wor1","word"]
Python Solution
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.