Introduction

In this blog post, we will explore how to solve the LeetCode problem "2273" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 result. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase using all the original letters exactly once. For example, "dacb" is an anagram of "abdc". Example 1: Input: words = ["abba","baba","bbaa","cd","cd"] Output: ["abba","cd"] Explanation: 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"] Explanation: 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.

Explanation

Here's the breakdown of the solution:

• Iterate and Compare: Process the words array sequentially, comparing each word with the previous one.
• Anagram Check: Efficiently determine if two words are anagrams by comparing their sorted forms.

• Conditional Appending: Only append a word to the result if it's not an anagram of the previous word in the result.

• Runtime Complexity: O(N * K log K), where N is the number of words and K is the maximum length of a word.

• Storage Complexity: O(N), where N is the number of words (in the worst case, where no anagrams are present).

Code

    def removeAnagrams(words):
    result = []
    last_sorted = ""  # Store the sorted form of the last word added to the result

    for word in words:
        sorted_word = "".join(sorted(word))  # Sort the word to check for anagrams
        if sorted_word != last_sorted:
            result.append(word)
            last_sorted = sorted_word

    return result