What is the time complexity of Encrypt and Decrypt Strings?

The optimal solution has a time complexity of O(N*M).

What is the space complexity of Encrypt and Decrypt Strings?

The space complexity is O(D*M).

What data structures are used to solve Encrypt and Decrypt Strings?

This problem uses Array, Hash Table, String, Design, Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Encrypt and Decrypt Strings?

Insight 1: Pre-compute and store encrypted words from the dictionary during initialization to optimize the `decrypt` function. This avoids repeated encryption calculations. Insight 2: Use a hash map (dictionary in Python, HashMap in Java, unordered_map in C++) to store the mappings between keys and values for efficient lookups during encryption.

Which companies ask LeetCode 2227. Encrypt and Decrypt Strings in interviews?

This problem is frequently asked at Duolingo.

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Encrypt and Decrypt Strings - LeetCode 2227 Solution

Encrypt and Decrypt Strings - Complete Solution Guide

Encrypt and Decrypt Strings is LeetCode problem 2227, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. You should implement a data structure that can encrypt or decrypt a 0-indexed string. A string is encrypted with the following process: For each character c in the string, we find the index i satisfying keys[i] == c in keys . Replace c with values[i] in the stri

Detailed Explanation

The problem requires designing a data structure that can encrypt and decrypt strings based on given keys, values, and a dictionary. The encryption process replaces characters in a string with corresponding 2-character strings from the 'values' array, based on their mapping in the 'keys' array. The decryption process reverses this, but with the possibility of multiple valid decryptions. The goal is to implement an `Encrypter` class with `encrypt` and `decrypt` methods. The `encrypt` method encrypts a given string. The `decrypt` method counts how many possible decryptions of a given string exist within a given 'dictionary'.

Solution Approach

The solution pre-computes the encrypted strings of all the dictionary words and stores their counts in a hash map (or similar data structure). During encryption, it iterates through the input string, replacing each character with its corresponding two-character value from the 'values' array, if a mapping exists. During decryption, it simply returns the number of times the input string appears in the pre-computed encrypted words.

Step-by-Step Algorithm

Step 1: Initialization: In the constructor, create a hash map to store the key-value mappings. Also, create another hash map to store the counts of encrypted dictionary words.
Step 2: Encryption Preparation: Iterate through the dictionary words. For each word, encrypt it using the key-value mappings. If the encrypted word is not empty (meaning the original word contains only chars present in keys), add the encrypted word to the hash map tracking the count of each word.
Step 3: Encryption: In the `encrypt` function, iterate through the input string. For each character, look up its corresponding value in the hash map. If the character is not found, return an empty string. Otherwise, append the corresponding value to the result string.
Step 4: Decryption: In the `decrypt` function, simply return the count of the input string from the encrypted words hash map. If the encrypted word is not found in the hashmap, return 0.

Key Insights

Insight 1: Pre-compute and store encrypted words from the dictionary during initialization to optimize the `decrypt` function. This avoids repeated encryption calculations.
Insight 2: Use a hash map (dictionary in Python, HashMap in Java, unordered_map in C++) to store the mappings between keys and values for efficient lookups during encryption.
Insight 3: The `decrypt` function needs to count the number of valid decryptions, but since the dictionary is fixed, we can simply encrypt the dictionary words and store the counts. This drastically improves the decryption operation.

Complexity Analysis

Time Complexity: O(N*M)

Space Complexity: O(D*M)

Topics

This problem involves: Array, Hash Table, String, Design, Trie.

Companies

Asked at: Duolingo.

Encrypt and Decrypt Strings | LeetCode 2227

Hard

Array Hash Table String Design Trie

Time: O(N*M)Space: O(D*M)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Encrypter {

    private Map<Character, String> keyToValue;
    private Map<String, Integer> encryptedDictCounts;

    public Encrypter(char[] keys, String[] values, String[] dictionary) {
        keyToValue = new HashMap<>();
        encryptedDictCounts = new HashMap<>();

        for (int i = 0; i < keys.length; i++) {
            keyToValue.put(keys[i], values[i]);
        }

        for (String word : dictionary) {
            String encryptedWord = encrypt(word);
            if (encryptedWord != null && !encryptedWord.isEmpty()) {
                encryptedDictCounts.put(encryptedWord, encryptedDictCounts.getOrDefault(encryptedWord, 0) + 1);
            }
        }
    }

    public String encrypt(String word1) {
        StringBuilder res = new StringBuilder();
        for (char char1 : word1.toCharArray()) {
            if (!keyToValue.containsKey(char1)) {
                return "";
            }
            res.append(keyToValue.get(char1));
        }
        return res.toString();
    }

    public int decrypt(String word2) {
        return encryptedDictCounts.getOrDefault(word2, 0);
    }
}

/**
 * Your Encrypter object will be instantiated and called as such:
 * Encrypter obj = new Encrypter(keys, values, dictionary);
 * String param_1 = obj.encrypt(word1);
 * int param_2 = obj.decrypt(word2);
 */

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Problem Info

Description

Problem 2227: Encrypt and Decrypt Strings - Detailed explanation and solution approaches.

Topics

Array Hash Table String Design Trie

Companies

Duolingo

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2227

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Initialization: In the constructor, create a hash map to store the key-value mappings. Also, create another hash map to store the counts of encrypted dictionary words.
2
Step 2: Encryption Preparation: Iterate through the dictionary words. For each word, encrypt it using the key-value mappings. If the encrypted word is not empty (meaning the original word contains only chars present in keys), add the encrypted word to the hash map tracking the count of each word.
3
Step 3: Encryption: In the `encrypt` function, iterate through the input string. For each character, look up its corresponding value in the hash map. If the character is not found, return an empty string. Otherwise, append the corresponding value to the result string.
4
Step 4: Decryption: In the `decrypt` function, simply return the count of the input string from the encrypted words hash map. If the encrypted word is not found in the hashmap, return 0.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N*M)

💾

Space Complexity

O(D*M)

📝

Why?

The constructor encrypts all words in dictionary to build the hashmap of encrypted strings and their counts. This operation takes O(N*M). The encrypt operation iterates over each character of word1 which is O(K). The decrypt operation simply checks the number of times word2 appears in the hashmap which is O(1). The space is dominated by the encrypted words in the dictionary.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Decryption with backtracking

An alternative approach is to perform the decryption process directly by trying all possible decryptions of the input string and checking if any of them exist in the dictionary. This could be implemented using recursion or backtracking.

Time: O(2^(N/2)), where N is the length of word2Space: O(N), where N is the length of word2

Trade-offs: This approach avoids pre-computation but can be much slower for larger dictionaries and longer input strings. It's generally less efficient than the pre-computation approach.

✨

Key Insights

1Insight 1: Pre-compute and store encrypted words from the dictionary during initialization to optimize the `decrypt` function. This avoids repeated encryption calculations.
2Insight 2: Use a hash map (dictionary in Python, HashMap in Java, unordered_map in C++) to store the mappings between keys and values for efficient lookups during encryption.
3Insight 3: The `decrypt` function needs to count the number of valid decryptions, but since the dictionary is fixed, we can simply encrypt the dictionary words and store the counts. This drastically improves the decryption operation.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: A character in the input to `encrypt` does not exist in `keys`. In this case, the `encrypt` function should return an empty string. The provided solutions handle this correctly by checking if the character exists in the `key_to_value` map/dictionary.
Edge case 2: The input `word2` to `decrypt` may not be a valid encrypted word derived from the given dictionary. The solutions handle this by returning 0 if the `word2` is not found in the `encrypted_dict_counts` map/dictionary.
Edge case 3: Empty dictionary. Should still work. No encrypted words means decrypt will always return 0. Handled correctly by initialization.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where a character in the input to `encrypt` is not found in the `keys`. Failing to return an empty string in this case would lead to incorrect encryption.

✗Common mistake 2: Incorrectly calculating the count of decrypted strings due to not properly caching encrypted dictionary words. This leads to repeatedly encrypting the dictionary words during decryption, which is inefficient.

✗Common mistake 3: Incorrectly calculating the space complexity by not accounting for the space used to store encrypted words. Each character gets replaced by a 2 char string leading to potentially double the space used.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data encryption where certain characters are always mapped to specific codes. This can be used to obfuscate data or prepare it for transmission over specific communication channels.

🏦Real-world scenario 2: Code generation where certain instructions are represented by specific character sequences. The encryption process simulates mapping high-level code to lower-level representations.

🎮Real-world scenario 3: Simplified version of a substitution cipher, used in cryptography.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Encrypt and Decrypt Strings"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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