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Commit deb93df

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Merge pull request #8 from 149ps/LC-problems
Lc problems
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"""
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You are given two strings of the same length s and t. In one step you can choose any character of t and replace it with another character.
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Return the minimum number of steps to make t an anagram of s.
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An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.
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Example 1:
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Input: s = "bab", t = "aba"
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Output: 1
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Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s.
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Example 2:
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Input: s = "leetcode", t = "practice"
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Output: 5
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Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s.
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Example 3:
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Input: s = "anagram", t = "mangaar"
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Output: 0
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Explanation: "anagram" and "mangaar" are anagrams.
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Constraints:
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1 <= s.length <= 5 * 104
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s.length == t.length
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s and t consist of lowercase English letters only.
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"""
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class Solution:
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def minSteps(self, s: str, t: str) -> int:
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counter_s = collections.Counter(s)
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counter_t = collections.Counter(t)
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return sum([val - counter_t[key] if val > counter_t[key] else 0 for key,val in counter_s.items()])
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"""
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Two strings are considered close if you can attain one from the other using the following operations:
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Operation 1: Swap any two existing characters.
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For example, abcde -> aecdb
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Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
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For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
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You can use the operations on either string as many times as necessary.
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Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.
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Example 1:
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Input: word1 = "abc", word2 = "bca"
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Output: true
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Explanation: You can attain word2 from word1 in 2 operations.
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Apply Operation 1: "abc" -> "acb"
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Apply Operation 1: "acb" -> "bca"
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Example 2:
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Input: word1 = "a", word2 = "aa"
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Output: false
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Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
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Example 3:
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Input: word1 = "cabbba", word2 = "abbccc"
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Output: true
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Explanation: You can attain word2 from word1 in 3 operations.
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Apply Operation 1: "cabbba" -> "caabbb"
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Apply Operation 2: "caabbb" -> "baaccc"
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Apply Operation 2: "baaccc" -> "abbccc"
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Constraints:
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1 <= word1.length, word2.length <= 105
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word1 and word2 contain only lowercase English letters.
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"""
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class Solution:
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def closeStrings(self, word1: str, word2: str) -> bool:
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count_word1 = collections.Counter(word1)
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count_word2 = collections.Counter(word2)
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counter1 = collections.Counter(count_word1.values())
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counter2 = collections.Counter(count_word2.values())
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return len(count_word1.keys() - count_word2.keys()) == 0 and len(counter1 - counter2) == 0

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