Return the minimum number of steps to make s and t anagrams of each other.

An anagram of a string is a string that contains the same characters with a different (or the same) ordering.

Example 1:

Input: s = "leetcode", t = "coats"
Output: 7
Explanation: 
- In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcodeas".
- In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coatsleede".
"leetcodeas" and "coatsleede" are now anagrams of each other.
We used a total of 2 + 5 = 7 steps.
It can be shown that there is no way to make them anagrams of each other with less than 7 steps.

Example 2:

Input: s = "night", t = "thing"
Output: 0
Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps.

Constraints:

• 1 <= s.length, t.length <= 2 * 105
• s and t consist of lowercase English letters.

Solution: Hashtable

Record the frequency difference of each letter.

Ans = sum(diff)

Time complexity: O(m + n)
Space complexity: O(26)

// Author: Huahua
class Solution {
public:
  int minSteps(string s, string t) {
    vector<int> diff(26);
    for (char c : s) ++diff[c - 'a'];
    for (char c : t) --diff[c - 'a'];
    int ans = 0;
    for (int d : diff)
      ans += abs(d);
    return ans;
  }
};

The post 花花酱 LeetCode 2186. Minimum Number of Steps to Make Two Strings Anagram II appeared first on Huahua's Tech Road.

Example:

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

Note:

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

Solution: HashTable

The sorted word will be the key of each group

Time complexity: O(sum(l*log(l)))
Space complexity: O(sum(l))

// Author: Huahua
class Solution {
public:
  vector<vector<string>> groupAnagrams(vector<string>& strs) {
    vector<vector<string>> ans;
    unordered_map<string, vector<int>> m;

    for (int i = 0; i < strs.size(); ++i) {
      string c = strs[i];
      sort(begin(c), end(c));
      m[c].push_back(i);
    }

    for (const auto& kv : m) {
      ans.push_back({});
      for (int i : kv.second)
        ans.back().push_back(strs[i]);    
    }

    return ans;
  }
};

The post 花花酱 LeetCode 49. Group Anagrams appeared first on Huahua's Tech Road.

Problem

题目大意：在s中找出所有p的Anagrams。

https://leetcode.com/problems/find-all-anagrams-in-a-string/description/

Given a string s and a non-empty string p, find all the start indices of p‘s anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

Solution: HashTable + Sliding Window

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 35 ms
class Solution {
public:
  vector<int> findAnagrams(string s, string p) {
    int n = s.length();
    int l = p.length();
    vector<int> ans;
    vector<int> vp(26, 0);
    vector<int> vs(26, 0);
    for (char c : p) ++vp[c - 'a'];    
    for (int i = 0; i < n; ++i) {
      if (i >= l) --vs[s[i - l] - 'a'];        
      ++vs[s[i] - 'a'];
      if (vs == vp) ans.push_back(i + 1 - l);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 438. Find All Anagrams in a String appeared first on Huahua's Tech Road.