#2273

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find Resultant Array After Removing Anagrams

This problem requires removing adjacent anagrams from a list of words using an array scanning and hash lookup approach.

Problem Statement

You are given a 0-indexed string array words, where words[i] consists of lowercase English letters. In one operation, you can remove an element from the array if it is an anagram of the previous element. The operation should be repeated as long as any two consecutive elements are anagrams. The task is to return the array after performing all operations.

The order of removal does not matter, as performing the operations in any order will yield the same result. Your goal is to return the resulting array with no adjacent anagrams.

Examples

Example 1

Input: words = ["abba","baba","bbaa","cd","cd"]

Output: ["abba","cd"]

One of the ways we can obtain the resultant array is by using the following operations:

Since words[2] = "bbaa" and words[1] = "baba" are anagrams, we choose index 2 and delete words[2]. Now words = ["abba","baba","cd","cd"].
Since words[1] = "baba" and words[0] = "abba" are anagrams, we choose index 1 and delete words[1]. Now words = ["abba","cd","cd"].
Since words[2] = "cd" and words[1] = "cd" are anagrams, we choose index 2 and delete words[2]. Now words = ["abba","cd"]. We can no longer perform any operations, so ["abba","cd"] is the final answer.

Example 2

Input: words = ["a","b","c","d","e"]

Output: ["a","b","c","d","e"]

No two adjacent strings in words are anagrams of each other, so no operations are performed.

Constraints

1 <= words.length <= 100
1 <= words[i].length <= 10
words[i] consists of lowercase English letters.

Solution Approach

Array Scanning with Hashmap

Scan through the array from left to right, and for each word, check if it is an anagram of the previous one by using a hash map or frequency counter. If the words are anagrams, remove the current word.

Sorting Approach

Sort the characters of each word and compare them. If two adjacent words have the same sorted version, they are anagrams, and the latter word can be removed.

Optimal Lookup Strategy

Instead of sorting the strings, use a frequency count to detect anagrams more efficiently. Compare the character frequencies of two adjacent words and remove the word if they match.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this problem depends on the method used. Sorting each word takes O(k log k) time, where k is the length of the word, and this operation is done for every word in the list. The space complexity depends on the approach, with the hash map method requiring O(k) space for each word's character frequency.

What Interviewers Usually Probe

Understanding of array scanning techniques and hash map usage.
Efficiency in detecting anagrams using frequency counts or sorting.
Ability to optimize for space and time complexity while solving.

Common Pitfalls or Variants

Common pitfalls

Failing to recognize the need for an efficient anagram check (sorting or hash map).
Not handling edge cases, like when no words are anagrams.
Overcomplicating the solution with unnecessary operations or data structures.

Follow-up variants

Allowing for a custom comparator instead of using frequency counts.
Modifying the input in place without using extra space.
Handling case sensitivity or extending the problem to a larger character set.

FAQ

What is the optimal way to check if two words are anagrams?

The most efficient way to check for anagrams is to use a frequency count of characters or to sort both strings and compare them.

How does this problem relate to the Array, Hash Table, and String topics?

This problem involves scanning through an array, using a hash table to store character frequencies or sorted versions of words, and dealing with strings in the process.

Can this problem be solved in-place?

Yes, it can be solved in-place by modifying the input array as you process the elements.

What is the time complexity of this problem?

The time complexity depends on the approach. Sorting each word would result in O(n * k log k), where n is the number of words and k is the length of the words. The hash map approach can be more efficient with a time complexity of O(n * k).

What are the common mistakes in solving this problem?

Common mistakes include failing to check for adjacent anagrams properly, overcomplicating the solution with unnecessary sorting, or not considering edge cases.

terminal

Solution

Solution 1: Simulation

We first add $\textit{words}[0]$ to the answer array, then traverse from $\textit{words}[1]$. If $\textit{words}[i - 1]$ and $\textit{words}[i]$ are not anagrams, we add $\textit{words}[i]$ to the answer array.

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class Solution:
    def removeAnagrams(self, words: List[str]) -> List[str]:
        def check(s: str, t: str) -> bool:
            if len(s) != len(t):
                return True
            cnt = Counter(s)
            for c in t:
                cnt[c] -= 1
                if cnt[c] < 0:
                    return True
            return False

        return [words[0]] + [t for s, t in pairwise(words) if check(s, t)]

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