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Posts tagged as “anagram”

花花酱 LeetCode 2186. Minimum Number of Steps to Make Two Strings Anagram II

By zxi on February 26, 2022

You are given two strings s and t. In one step, you can append any character to either s or t.

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)

C++

花花酱 LeetCode 49. Group Anagrams

By zxi on October 2, 2019

Given an array of strings, group anagrams together.

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))

C++

花花酱 LeetCode 438. Find All Anagrams in a String

By zxi on May 28, 2018

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++