What is the time complexity of Verifying an Alien Dictionary?

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

What data structures are used to solve Verifying an Alien Dictionary?

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

What are the key insights for solving Verifying an Alien Dictionary?

Insight 1: The core idea is to create a mapping between the alien alphabet and the standard alphabet to easily compare characters. Using a dictionary or array to store the index of each letter in the `order` string is crucial. Insight 2: Comparing two words involves iterating through their characters until a difference is found or one word runs out of characters. If one word is a prefix of another, the shorter word must come first in lexicographical order.

Which companies ask LeetCode 953. Verifying an Alien Dictionary in interviews?

This problem is frequently asked at Snap, Wix.

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Verifying an Alien Dictionary - LeetCode 953 Solution

Verifying an Alien Dictionary - Complete Solution Guide

Verifying an Alien Dictionary is LeetCode problem 953, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In an alien language, surprisingly, they also use English lowercase letters, but possibly in a different order . The order of the alphabet is some permutation of lowercase letters. Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographically in this alien language. Example 1: Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz" Output: true Explanation: As 'h' comes before 'l'

Detailed Explanation

The problem asks us to determine if a given list of words (`words`) is sorted according to a custom alien alphabet (`order`). The `order` string defines the order of the lowercase English letters. We need to compare the words in the `words` list lexicographically, using the alien alphabet defined by `order`, and return `true` if the words are sorted correctly, and `false` otherwise. The words only contain lowercase English letters, and the order string is a permutation of all 26 lowercase letters.

Solution Approach

The solution involves creating a mapping (using a dictionary or array) from each character in the `order` string to its index. This mapping represents the alien alphabet's ordering. Then, we iterate through the `words` list, comparing adjacent pairs of words using the alien alphabet's ordering. If any adjacent pair is out of order, we return `false`. If we reach the end of the list without finding any out-of-order pairs, we return `true`.

Step-by-Step Algorithm

Step 1: Create a map (dictionary or array) called `order_map` that stores the index of each character in the `order` string. For example, if `order` is "hlabcdefgijkmnopqrstuvwxyz", then `order_map['h']` will be 0, `order_map['l']` will be 1, `order_map['a']` will be 2, and so on.
Step 2: Iterate through the `words` list from index 0 to `len(words) - 2` (inclusive), comparing each word with the next word.
Step 3: For each pair of words (word1, word2), compare their characters one by one until you find a difference or reach the end of one of the words.
Step 4: If a character difference is found, compare their indices in `order_map`. If `order_map[word1[i]] > order_map[word2[i]]`, then word1 comes after word2, so the words are out of order. Return `false`.
Step 5: If one word is a prefix of the other (i.e., all characters are the same up to the length of the shorter word), the shorter word must come first. If word1 is longer than word2, the words are out of order. Return `false`.
Step 6: If the loop completes without finding any out-of-order pairs, the words are sorted according to the alien alphabet. Return `true`.

Key Insights

Insight 1: The core idea is to create a mapping between the alien alphabet and the standard alphabet to easily compare characters. Using a dictionary or array to store the index of each letter in the `order` string is crucial.
Insight 2: Comparing two words involves iterating through their characters until a difference is found or one word runs out of characters. If one word is a prefix of another, the shorter word must come first in lexicographical order.
Insight 3: The most efficient solution checks the order of adjacent word pairs, returning immediately if an unsorted pair is encountered.

Complexity Analysis

Time Complexity: O(N*M)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Snap, Wix.

Verifying an Alien Dictionary | LeetCode 953

Easy

Array Hash Table String

Time: O(N*M)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean isAlienSorted(String[] words, String order) {
        int[] charMap = new int[26];
        for (int i = 0; i < order.length(); i++) {
            charMap[order.charAt(i) - 'a'] = i;
        }

        for (int i = 0; i < words.length - 1; i++) {
            String word1 = words[i];
            String word2 = words[i + 1];
            int minLength = Math.min(word1.length(), word2.length());

            boolean foundDiff = false;
            for (int j = 0; j < minLength; j++) {
                char char1 = word1.charAt(j);
                char char2 = word2.charAt(j);

                if (char1 != char2) {
                    if (charMap[char1 - 'a'] > charMap[char2 - 'a']) {
                        return false;
                    }
                    foundDiff = true;
                    break;
                }
            }

            if (!foundDiff && word1.length() > word2.length()) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 953: Verifying an Alien Dictionary - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

SnapWix

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:953

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

6 steps

1
Step 1: Create a map (dictionary or array) called `order_map` that stores the index of each character in the `order` string. For example, if `order` is "hlabcdefgijkmnopqrstuvwxyz", then `order_map['h']` will be 0, `order_map['l']` will be 1, `order_map['a']` will be 2, and so on.
2
Step 2: Iterate through the `words` list from index 0 to `len(words) - 2` (inclusive), comparing each word with the next word.
3
Step 3: For each pair of words (word1, word2), compare their characters one by one until you find a difference or reach the end of one of the words.
4
Step 4: If a character difference is found, compare their indices in `order_map`. If `order_map[word1[i]] > order_map[word2[i]]`, then word1 comes after word2, so the words are out of order. Return `false`.
5
Step 5: If one word is a prefix of the other (i.e., all characters are the same up to the length of the shorter word), the shorter word must come first. If word1 is longer than word2, the words are out of order. Return `false`.
6
Step 6: If the loop completes without finding any out-of-order pairs, the words are sorted according to the alien alphabet. Return `true`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the nested loops structure. The outer loop iterates through the `words` list which has length `N`. The inner loop compares characters in two words, at worst iterating up to the length of the longest word which is bounded by `M`. This leads to O(N*M). The space complexity is O(1) because we use a fixed-size map (26 entries) to store the character order. The size of this map doesn't depend on the input size. There are no significant auxiliary data structures.

📖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

Using functools.cmp_to_key in Python

In Python, one can use `functools.cmp_to_key` to create a key function based on a custom comparison function. This key function can then be used with the `sorted()` function to sort the words according to the alien alphabet. Then you compare if the sorted result is the same as original words.

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

Trade-offs: This approach can be more concise in Python but is typically slower because sorting has a logarithmic factor (logN) and the comparison is done with cost M, and uses additional memory O(N) for sorting.

✨

Key Insights

1Insight 1: The core idea is to create a mapping between the alien alphabet and the standard alphabet to easily compare characters. Using a dictionary or array to store the index of each letter in the `order` string is crucial.
2Insight 2: Comparing two words involves iterating through their characters until a difference is found or one word runs out of characters. If one word is a prefix of another, the shorter word must come first in lexicographical order.
3Insight 3: The most efficient solution checks the order of adjacent word pairs, returning immediately if an unsorted pair is encountered.

📚

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: `words` is empty or has only one word. In this case, the words are considered sorted, and the algorithm should return `true`. The code implicitly handles this as the for loop comparing words will not execute if there are less than 2 words.
Edge case 2: One word is a prefix of another. For example, `words = ["apple", "app"]`. The longer word ("apple") should come after the shorter word ("app"). The algorithm correctly handles this by checking the lengths of the words after comparing their characters up to the length of the shorter word.
Edge case 3: Duplicate words. For example, `words = ["apple", "apple"]`. The algorithm should return `true` because duplicate words are considered to be sorted.
Edge case 4: The `order` string is not a valid permutation of lowercase English letters. The problem states this is a constraint, so the code doesn't need to handle it, but a more robust solution might include validation of the input `order` string.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where one word is a prefix of another. For instance, assuming that if all characters are the same, the words are sorted, without considering the length difference.

✗Common mistake 2: Incorrectly comparing characters by directly using their ASCII values instead of looking up their indices in the `order_map`.

✗Common mistake 3: Not handling empty inputs or edge cases where the number of words is less than 2. This can cause the program to crash or return incorrect results.

✅

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: Custom sorting in applications that require user-defined orderings, such as sorting files by custom metadata or sorting database results based on user preferences.

🏦Real-world scenario 2: Language processing tasks where different languages or dialects have different alphabets or character orderings. For example, sorting strings in a language with diacritics or a different collation order.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: July 24, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Verifying an Alien Dictionary"?

For this easy-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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