Reference: LeetCode
Difficulty: Medium

Problem

Given an array of strings, group anagrams together.

Note:

  • All inputs will be in lowercase.
  • The order of your output does not matter.

Example: 

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

Analysis

HashMap + Sorting

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// output order does not matter
// inputs are in lowercase
public List<List<String>> groupAnagrams(String[] strs) {
  if (strs == null || strs.length == 0) {
    return new ArrayList<>();
  }
  Map<String, List<String>> map = new HashMap<>();
  for (String s : strs) {
    String key = sortString(s);
    List<String> list = map.getOrDefault(key, new ArrayList<>());
    list.add(s);
    map.put(key, list);
  }
  // List<List<String>> result = new ArrayList<>();
  // for (String key : map.keySet()) {
    // result.add(map.get(key));
  // }
  // return result;
  return new ArrayList<>(map.values());
}

private String sortString(String str) {
  char[] arr = str.toCharArray();
  Arrays.sort(arr);
  return String.valueOf(arr);
}

Time: $O(NK\log{K})$ where $K$ is the maximum length of a string.
Space: $O(NK)$

Categorize by Count

Count a string’s characters and generate a key like #1#2#3#4....

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public List<List<String>> groupAnagrams(String[] strs) {
  if (strs.length == 0) return new ArrayList();
  Map<String, List> ans = new HashMap<String, List>();
  int[] count = new int[26];
  for (String s : strs) {
    Arrays.fill(count, 0);
    for (char c : s.toCharArray()) count[c - 'a']++;

    StringBuilder sb = new StringBuilder("");
    for (int i = 0; i < 26; i++) {
      sb.append('#');
      sb.append(count[i]);
    }
    String key = sb.toString();
    if (!ans.containsKey(key)) ans.put(key, new ArrayList());
    ans.get(key).add(s);
  }
  return new ArrayList(ans.values());
}

Time: $O(NK)$
Space: $O(NK)$