Leetcode #820: Short Encoding of Words
In this guide, we solve Leetcode #820 Short Encoding of Words 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 valid encoding of an array of words is any reference string s and array of indices indices such that: words.length == indices.length The reference string s ends with the '#' character. For each index indices[i], the substring of s starting from indices[i] and up to (but not including) the next '#' character is equal to words[i].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Trie, Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: words = ["time", "me", "bell"]
Output: 10
Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"
Python Solution
class Trie:
def __init__(self) -> None:
self.children = [None] * 26
class Solution:
def minimumLengthEncoding(self, words: List[str]) -> int:
root = Trie()
for w in words:
cur = root
for c in w[::-1]:
idx = ord(c) - ord("a")
if cur.children[idx] == None:
cur.children[idx] = Trie()
cur = cur.children[idx]
return self.dfs(root, 1)
def dfs(self, cur: Trie, l: int) -> int:
isLeaf, ans = True, 0
for i in range(26):
if cur.children[i] != None:
isLeaf = False
ans += self.dfs(cur.children[i], l + 1)
if isLeaf:
ans += l
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
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.