Group Anagrams LeetCode Solution | Easy Approach

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Group Anagrams | Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

Constraints:

  • 1 <= strs.length <= 104
  • 0 <= strs[i].length <= 100
  • strs[i] consists of lowercase English letters.

Group Anagrams Solutions

Time: O(n)
Space: O(nk)

C++

 class Solution {
 public:
  vector<vector<string>> groupAnagrams(vector<string>& strs) {
    vector<vector<string>> ans;
    unordered_map<string, vector<string>> keyToAnagrams;

    for (const string& str : strs) {
      string key = str;
      sort(begin(key), end(key));
      keyToAnagrams[key].push_back(str);
    }

    for (const auto& [_, anagrams] : keyToAnagrams)
      ans.push_back(anagrams);

    return ans;
  }
};

Java

 class Solution {
  public List<List<String>> groupAnagrams(String[] strs) {
    Map<String, List<String>> keyToAnagrams = new HashMap<>();

    for (final String str : strs) {
      char[] chars = str.toCharArray();
      Arrays.sort(chars);
      String key = String.valueOf(chars);
      keyToAnagrams.computeIfAbsent(key, k -> new ArrayList<>()).add(str);
    }

    return new ArrayList<>(keyToAnagrams.values());
  }
}

Python

 class Solution:
  def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
    dict = defaultdict(list)

    for str in strs:
      key = ''.join(sorted(str))
      dict[key].append(str)

    return dict.values()

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