LeetCode 2273: Find Resultant Array After Removing Anagrams ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2026-01-29

Submissions: Accepted (first try)

Approach
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After deleting words[i], the next word’s “previous” becomes words[i - 1]. The deleted word was an anagram of words[i - 1], and anagrams are transitive, so comparing with either gives the same result.

So scan left to right and compare only with the last kept word: keep it if not an anagram. Check anagrams with sorted strings (Hash Table Template 5).

My Solution
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I originally went right to left, comparing neighbors with Counter and popping when equal. Here it builds a new list left to right:

class Solution:
    def removeAnagrams(self, words: List[str]) -> List[str]:
        res = []
        for w in words:
            if not res or sorted(res[-1]) != sorted(w):
                res.append(w)
        return res

Complexity
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Time O(n · L log L) for word length L, space O(n · L).

Mistakes to Avoid
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pop inside a loop shifts indices; it is only safe right to left, and building a new list is simpler.

Interview Explanation
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After a deletion the next word compares with the one before the deleted word; anagrams are transitive, so that equals comparing with the last kept word. I scan left to right and keep a word when its sorted form differs from the last kept one. Time O(n L log L).