Leetcode #2227: Encrypt and Decrypt Strings
In this guide, we solve Leetcode #2227 Encrypt and Decrypt Strings 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
You are given a character array keys containing unique characters and a string array values containing strings of length 2. You are also given another string array dictionary that contains all permitted original strings after decryption.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, 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
["Encrypter", "encrypt", "decrypt"]
[[['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]], ["abcd"], ["eizfeiam"]]
Output
[null, "eizfeiam", 2]
Explanation
Encrypter encrypter = new Encrypter([['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]);
encrypter.encrypt("abcd"); // return "eizfeiam".
// 'a' maps to "ei", 'b' maps to "zf", 'c' maps to "ei", and 'd' maps to "am".
encrypter.decrypt("eizfeiam"); // return 2.
// "ei" can map to 'a' or 'c', "zf" maps to 'b', and "am" maps to 'd'.
// Thus, the possible strings after decryption are "abad", "cbad", "abcd", and "cbcd".
// 2 of those strings, "abad" and "abcd", appear in dictionary, so the answer is 2.
Python Solution
class Encrypter:
def __init__(self, keys: List[str], values: List[str], dictionary: List[str]):
self.mp = dict(zip(keys, values))
self.cnt = Counter(self.encrypt(v) for v in dictionary)
def encrypt(self, word1: str) -> str:
res = []
for c in word1:
if c not in self.mp:
return ''
res.append(self.mp[c])
return ''.join(res)
def decrypt(self, word2: str) -> int:
return self.cnt[word2]
# Your Encrypter object will be instantiated and called as such:
# obj = Encrypter(keys, values, dictionary)
# param_1 = obj.encrypt(word1)
# param_2 = obj.decrypt(word2)
Complexity
The time complexity is , where and are the lengths of and , respectively. 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.