How do you solve Find Resultant Array After Removing Anagrams on LeetCode?

Find Resultant Array After Removing Anagrams (LeetCode #2273) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k log k) time complexity and O(n) space complexity. Key insight: Understanding anagrams requires recognizing that they have the same characters in different orders.

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

The optimal solution for Find Resultant Array After Removing Anagrams runs in O(n * k log k) time and uses O(n) space. Here, n is the number of words and k is the average length of the words. Sorting each word takes O(k log k), and we store each unique sorted word in a HashSet.

What is the brute force solution for Find Resultant Array After Removing Anagrams?

The brute force approach for Find Resultant Array After Removing Anagrams is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each pair of adjacent words to see if they are anagrams. If they are, we remove one of them and repeat until no more anagrams are found. The optimal Optimal Solution approach improves this to O(n * k log k).

What algorithm or data structure is used to solve Find Resultant Array After Removing Anagrams?

Find Resultant Array After Removing Anagrams uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Resultant Array After Removing Anagrams?

Always clarify the problem and ask about edge cases before jumping into coding. Explain your thought process as you code; it shows your understanding. Consider both time and space complexity when discussing your approach.

What are common mistakes when solving Find Resultant Array After Removing Anagrams?

Not considering edge cases like empty strings or single-character words. Failing to use a data structure like a HashSet to track seen words.

#2273

Find Resultant Array After Removing Anagrams

Easy

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k log k)
Space	O(1)	O(n)

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Intuition

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

The optimal approach uses a HashSet to track unique sorted versions of words. This allows us to efficiently determine if a word is an anagram of any previously seen word.

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Algorithm

3 steps

1Step 1: Initialize an empty list for the result and a HashSet to track seen sorted words.
2Step 2: Iterate through each word in the input list.
3Step 3: Sort the current word and check if it exists in the HashSet. If it does, skip it; if not, add it to both the HashSet and the result list.

solution.py10 lines

1# Full working Python code
2words = ["abba", "baba", "bbaa", "cd", "cd"]
3result = []
4seen = set()
5for word in words:
6    sorted_word = ''.join(sorted(word))
7    if sorted_word not in seen:
8        seen.add(sorted_word)
9        result.append(word)
10print(result)

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Complexity note: Here, n is the number of words and k is the average length of the words. Sorting each word takes O(k log k), and we store each unique sorted word in a HashSet.

1Understanding anagrams requires recognizing that they have the same characters in different orders.
2Using sorting or frequency counting can help identify anagrams efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.