LeetCode 2273: Find Resultant Array After Removing Anagrams
LeetCode 2273 Solution Explanation
Explanation:
To solve this problem, we can iterate through the array of words and check if the current word and the previous word are anagrams of each other. If they are anagrams, we remove the current word from the array. We repeat this process until we can no longer find anagrams adjacent to each other.
This can be done efficiently by using a HashSet to store the sorted characters of each word. When comparing two words for anagrams, we can check if their sorted character sets are equal in O(1) time.
Algorithm:
- Initialize a HashSet to store the sorted characters of a word.
- Iterate through the array of words.
- For each word, convert it to a char array, sort the characters, and convert back to a string.
- Check if the sorted characters are already in the HashSet. If yes, remove the word from the array.
- If not, add the sorted characters to the HashSet.
- Continue until no more anagrams can be removed.
Time Complexity: O(n * k * log(k)), where n is the number of words and k is the maximum length of a word.
Space Complexity: O(n * k), where n is the number of words and k is the maximum length of a word.
LeetCode 2273 Solutions in Java, C++, Python
import java.util.*;
class Solution {
public List<String> findResultantArray(String[] words) {
Set<String> set = new HashSet<>();
List<String> result = new ArrayList<>();
for (String word : words) {
char[] chars = word.toCharArray();
Arrays.sort(chars);
String sortedWord = new String(chars);
if (set.contains(sortedWord)) {
// Already encountered an anagram, skip this word
} else {
set.add(sortedWord);
result.add(word);
}
}
return result;
}
}Interactive Code Editor for LeetCode 2273
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