2273. Find Resultant Array After Removing Anagrams
Description
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.
Solutions
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.
The problem is converted to determining whether two strings are anagrams. We define a helper function $\textit{check}(s, t)$ to achieve this. If $s$ and $t$ are not anagrams, we return $\text{true}$; otherwise, we return $\text{false}$.
In the function $\textit{check}(s, t)$, we first check if the lengths of $s$ and $t$ are equal. If they are not, we return $\text{true}$. Otherwise, we use an array $\textit{cnt}$ of length $26$ to count the occurrences of each character in $s$, then traverse each character in $t$ and decrement $\textit{cnt}[c]$ by $1$. If $\textit{cnt}[c]$ is less than $0$, we return $\text{true}$. If we traverse all characters in $t$ without issues, it means $s$ and $t$ are anagrams, and we return $\text{false}$.
The time complexity is $O(L)$, and the space complexity is $O(|\Sigma|)$. Here, $L$ is the length of the array $\textit{words}$, and $\Sigma$ is the character set, which is lowercase English letters, so $|\Sigma| = 26$.
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