What is the time complexity of Find Resultant Array After Removing Anagrams?

The optimal solution has a time complexity of O(n*k*log(k)).

What is the space complexity of Find Resultant Array After Removing Anagrams?

The space complexity is O(n*k).

What data structures are used to solve Find Resultant Array After Removing Anagrams?

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

What are the key insights for solving Find Resultant Array After Removing Anagrams?

Insight 1: Anagrams have the same character frequencies. We can sort the characters of each string to easily compare if they are anagrams. Insight 2: A single pass through the array is sufficient. We only need to compare each word with the previous word in the resulting array.

Which companies ask LeetCode 2273. Find Resultant Array After Removing Anagrams in interviews?

This problem is frequently asked at J.P. Morgan.

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Find Resultant Array After Removing Anagrams - LeetCode 2273 Solution

Q: How to solve Find Resultant Array After Removing Anagrams?

The solutions all utilize a similar approach. They iterate through the input array, comparing each word to the last word added to the result array. The comparison is done by sorting the characters of the words. If the sorted versions are different (meaning they are not anagrams), the current word is appended to the result array. This is a greedy approach as it makes a locally optimal choice at each step (adding a word only if it is not an anagram of the previous word). The greedy approach works correctly here because the order of removals does not affect the final outcome.

Find Resultant Array After Removing Anagrams - Complete Solution Guide

Find Resultant Array After Removing Anagrams is LeetCode problem 2273, 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

You are given a 0-indexed string array words , where words[i] consists of lowercase English letters. In one operation, select any index i such that 0 < i < words.length and words[i - 1] and words[i] are anagrams , and delete words[i] from words . Keep performing this operation as long as you can select an index that satisfies the conditions. Return words after performing all operations . It can be shown that selecting the indices for each operation in any arbitrary order will lead to the same re

Detailed Explanation

The problem asks you to remove words from a list of strings (words) if they are anagrams of the preceding word. An anagram is a word formed by rearranging the letters of another word. The input is a 0-indexed array of strings. The output is the modified array after removing all such anagrams. The order of remaining words should be preserved. You can perform operations in any order and still get the same final result. For instance, if you have "abba", "baba", "bbaa", they are all anagrams of each other and only one should remain.

Solution Approach

The solutions all utilize a similar approach. They iterate through the input array, comparing each word to the last word added to the result array. The comparison is done by sorting the characters of the words. If the sorted versions are different (meaning they are not anagrams), the current word is appended to the result array. This is a greedy approach as it makes a locally optimal choice at each step (adding a word only if it is not an anagram of the previous word). The greedy approach works correctly here because the order of removals does not affect the final outcome.

Step-by-Step Algorithm

Step 1: Initialize an empty result array (list, vector, etc.).
Step 2: Add the first word from the input array to the result array.
Step 3: Iterate through the remaining words in the input array, starting from the second word.
Step 4: For each word, sort its characters.
Step 5: Compare the sorted current word with the sorted last word in the result array.
Step 6: If the sorted words are different (not anagrams), add the current word to the result array.
Step 7: Return the result array.

Key Insights

Insight 1: Anagrams have the same character frequencies. We can sort the characters of each string to easily compare if they are anagrams.
Insight 2: A single pass through the array is sufficient. We only need to compare each word with the previous word in the resulting array.
Insight 3: Using a simple list (or vector/array list) to store the resultant array and efficiently appending strings based on the anagram check optimizes space usage compared to using more complex data structures.

Complexity Analysis

Time Complexity: O(n*k*log(k))

Space Complexity: O(n*k)

Topics

This problem involves: Array, Hash Table, String, Sorting.

Companies

Asked at: J.P. Morgan.

Find Resultant Array After Removing Anagrams | LeetCode 2273

Easy

Array Hash Table String Sorting

Time: O(n*k*log(k))Space: O(n*k)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<String> removeAnagrams(String[] words) {
        List<String> result = new ArrayList<>();
        if (words == null || words.length == 0) {
            return result;
        }

        result.add(words[0]);
        for (int i = 1; i < words.length; i++) {
            if (!isAnagram(words[i - 1], words[i])) {
                result.add(words[i]);
            }
        }

        return result;
    }

    private boolean isAnagram(String s1, String s2) {
        if (s1.length() != s2.length()) {
            return false;
        }

        char[] arr1 = s1.toCharArray();
        char[] arr2 = s2.toCharArray();
        Arrays.sort(arr1);
        Arrays.sort(arr2);

        return Arrays.equals(arr1, arr2);
    }
}

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Problem Info

Description

Problem 2273: Find Resultant Array After Removing Anagrams - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting

Companies

J.P. Morgan

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2273

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

7 steps

1
Step 1: Initialize an empty result array (list, vector, etc.).
2
Step 2: Add the first word from the input array to the result array.
3
Step 3: Iterate through the remaining words in the input array, starting from the second word.
4
Step 4: For each word, sort its characters.
5
Step 5: Compare the sorted current word with the sorted last word in the result array.
6
Step 6: If the sorted words are different (not anagrams), add the current word to the result array.
7
Step 7: Return the result array.

💡

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*k*log(k))

💾

Space Complexity

O(n*k)

📝

Why?

The dominant operation is the sorting of characters within each word. Sorting k characters takes O(k * log k) time. This is performed for each of the n words. The space complexity comes from storing the resulting array of words, which in the worst case contains all n words. Each word has length k, giving a space complexity of O(n * k).

📖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 Character Counts

Instead of sorting, you can create a frequency count array (26 elements for lowercase English letters) for each word and compare those arrays. Two words are anagrams if and only if their frequency count arrays are identical.

Time: O(n * k)Space: O(n * k)

Trade-offs: This approach has a better time complexity if you consider the average case performance of sorting algorithms. In the average case, counting character frequency will be faster. However, the space complexity remains similar. The sorting approach is often preferred for simplicity and readability.

✨

Key Insights

1Insight 1: Anagrams have the same character frequencies. We can sort the characters of each string to easily compare if they are anagrams.
2Insight 2: A single pass through the array is sufficient. We only need to compare each word with the previous word in the resulting array.
3Insight 3: Using a simple list (or vector/array list) to store the resultant array and efficiently appending strings based on the anagram check optimizes space usage compared to using more complex data structures.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The solutions handle this by returning an empty array.
Edge case 2: All words are anagrams of each other. Only the first word will be preserved in the output.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the characters before comparing. Direct string comparison won't work for anagrams.

✗Common mistake 2: Inefficient comparison of anagrams. Using nested loops for comparing character frequencies without an optimized technique can lead to O(n*k^2) complexity.

✅

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: Text processing and data cleaning. Removing duplicate words (considered as anagrams) to reduce redundancy in a document.

🏦Real-world scenario 2: Security applications. Detecting potential anagram-based passwords or usernames to enhance security systems.

💡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: September 22, 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 "Find Resultant Array After Removing Anagrams"?

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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