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# Find Resultant Array After Removing Anagrams LeetCode Solution

## Problem – Find Resultant Array After Removing Anagrams LeetCode Solution

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 = "bbaa" and words = "baba" are anagrams, we choose index 2 and delete words.
Now words = ["abba","baba","cd","cd"].
- Since words = "baba" and words = "abba" are anagrams, we choose index 1 and delete words.
Now words = ["abba","cd","cd"].
- Since words = "cd" and words = "cd" are anagrams, we choose index 2 and delete words.
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.

### Find Resultant Array After Removing Anagrams LeetCode Solution in C++

``````vector<string> removeAnagrams(vector<string>& A) {
vector<string> ans;
int n= size(A);
for(int i=0;i<n;){
int j=i+1;
while( j<n and isang(A[i],A[j]) ) j++;
ans.push_back(A[i]);
i=j;
}
return ans;
}
//function for checking if two string are anagrams or not
bool isang(string p,string q){
vector<int> cnt(26);
int res=0;
for(auto i:p) cnt[i-'a']++;
for(auto i:q) cnt[i-'a']--;
for(auto i:cnt) if(i!=0) return false;
return true;
}``````

### Find Resultant Array After Removing Anagrams LeetCode Solution in Java

``````public List<String> removeAnagrams(String[] A) {
List<String> ans= new ArrayList<>();
int n= A.length;
for(int i=0;i<n;){
int j=i+1;
while( j<n && isang(A[i],A[j]) ) j++;
i=j;
}
return ans;
}
//function for checking if two string are anagrams or not
boolean isang(String p,String q){
int[] cnt= new int;
int res=0;
for(int i=0;i<p.length();i++) cnt[p.charAt(i)-'a']++;
for(int i=0;i<q.length();i++) cnt[q.charAt(i)-'a']--;
for(int i:cnt) if(i!=0) return false;
return true;
}``````

### Find Resultant Array After Removing Anagrams LeetCode Solution in Python

``````class Solution:
def removeAnagrams(self, words: List[str]) -> List[str]:
if len(words)==1:
return words
i = 1
while i<len(words):
if sorted(list(words[i]))==sorted(list(words[i-1])):
words.pop(i)
else:
i+=1
return words``````
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