2273. Find Resultant Array After Removing Anagrams LeetCode Solution
In this guide, you will get 2273. Find Resultant Array After Removing Anagrams LeetCode Solution with the best time and space complexity. The solution to Find Resultant Array After Removing Anagrams problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Find Resultant Array After Removing Anagrams
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.
Complexity Analysis
Time Complexity: O(\Sigma |\texttt{words[i]}|)
Space Complexity: O(\Sigma |\texttt{words[i]}|)
2273. Find Resultant Array After Removing Anagrams LeetCode Solution in C++
class Solution {
public:
vector<string> removeAnagrams(vector<string>& words) {
vector<string> ans;
for (int i = 0; i < words.size();) {
int j = i + 1;
while (j < words.size() && isAnagram(words[i], words[j]))
++j;
ans.push_back(words[i]);
i = j;
}
return ans;
}
private:
bool isAnagram(const string& a, const string& b) {
if (a.length() != b.length())
return false;
vector<int> count(26);
for (const char c : a)
++count[c - 'a'];
for (const char c : b)
--count[c - 'a'];
return ranges::all_of(count, [](const int c) { return c == 0; });
}
}; /* code provided by PROGIEZ */
2273. Find Resultant Array After Removing Anagrams LeetCode Solution in Java
class Solution {
public List<String> removeAnagrams(String[] words) {
List<String> ans = new ArrayList<>();
for (int i = 0; i < words.length;) {
int j = i + 1;
while (j < words.length && isAnagram(words[i], words[j]))
++j;
ans.add(words[i]);
i = j;
}
return ans;
}
private boolean isAnagram(final String a, final String b) {
if (a.length() != b.length())
return false;
int[] count = new int[26];
for (final char c : a.toCharArray())
++count[c - 'a'];
for (final char c : b.toCharArray())
--count[c - 'a'];
return Arrays.stream(count).allMatch(c -> c == 0);
}
} // code provided by PROGIEZ
2273. Find Resultant Array After Removing Anagrams LeetCode Solution in Python
class Solution:
def removeAnagrams(self, words: list[str]) -> list[str]:
ans = []
def isAnagram(a: str, b: str) -> bool:
count = collections.Counter(a)
count.subtract(collections.Counter(b))
return all(value == 0 for value in count.values())
i = 0
while i < len(words):
j = i + 1
while j < len(words) and isAnagram(words[i], words[j]):
j += 1
ans.append(words[i])
i = j
return ans # code by PROGIEZ
