Lc problems

1347. Minimum Number of Steps to Make Two Strings Anagram/1347. Minimum Number of Steps to Make Two Strings Anagram.py

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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.
+
+Return the minimum number of steps to make t an anagram of s.
+
+An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.
+
+
+
+Example 1:
+
+Input: s = "bab", t = "aba"
+Output: 1
+Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s.
+Example 2:
+
+Input: s = "leetcode", t = "practice"
+Output: 5
+Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s.
+Example 3:
+
+Input: s = "anagram", t = "mangaar"
+Output: 0
+Explanation: "anagram" and "mangaar" are anagrams.
+
+
+Constraints:
+
+1 <= s.length <= 5 * 104
+s.length == t.length
+s and t consist of lowercase English letters only.
+"""
+class Solution:
+    def minSteps(self, s: str, t: str) -> int:
+        counter_s = collections.Counter(s)
+        counter_t = collections.Counter(t)
+        return sum([val - counter_t[key] if val > counter_t[key] else 0 for key,val in counter_s.items()])

1657. Determine if Two Strings Are Close/1657. Determine if Two Strings Are Close.py

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+"""
+Two strings are considered close if you can attain one from the other using the following operations:
+
+Operation 1: Swap any two existing characters.
+For example, abcde -> aecdb
+Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
+For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
+You can use the operations on either string as many times as necessary.
+
+Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.
+
+
+
+Example 1:
+
+Input: word1 = "abc", word2 = "bca"
+Output: true
+Explanation: You can attain word2 from word1 in 2 operations.
+Apply Operation 1: "abc" -> "acb"
+Apply Operation 1: "acb" -> "bca"
+Example 2:
+
+Input: word1 = "a", word2 = "aa"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
+Example 3:
+
+Input: word1 = "cabbba", word2 = "abbccc"
+Output: true
+Explanation: You can attain word2 from word1 in 3 operations.
+Apply Operation 1: "cabbba" -> "caabbb"
+Apply Operation 2: "caabbb" -> "baaccc"
+Apply Operation 2: "baaccc" -> "abbccc"
+
+
+Constraints:
+
+1 <= word1.length, word2.length <= 105
+word1 and word2 contain only lowercase English letters.
+"""
+class Solution:
+    def closeStrings(self, word1: str, word2: str) -> bool:
+        count_word1 = collections.Counter(word1)
+        count_word2 = collections.Counter(word2)
+        counter1 = collections.Counter(count_word1.values())
+        counter2 = collections.Counter(count_word2.values())
+        return len(count_word1.keys() - count_word2.keys()) == 0  and len(counter1 - counter2) == 0