Determine if Two Strings Are Close

Author: anantkaushik

1657-determine-if-two-strings-are-close.py

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+"""
+Problem Link: https://leetcode.com/problems/determine-if-two-strings-are-close/
+
+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"
+
+Example 4:
+Input: word1 = "cabbba", word2 = "aabbss"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any amount of operations.
+ 
+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:
+ letters1 = [0] * 26
+ letters2 = [0] * 26
+ for word in word1:
+ letters1[ord(word) - ord('a')] += 1
+ 
+ for word in word2:
+ letters2[ord(word) - ord('a')] += 1
+ 
+ 
+ count = {}
+ for i in range(len(letters1)):
+ if letters1[i]:
+ if letters1[i] in count and letters1[count[letters1[i]]]:
+ count.pop(letters1[i])
+ else:
+ count[letters1[i]] = i
+ if letters2[i]:
+ if letters2[i] in count and letters2[count[letters2[i]]]:
+ count.pop(letters2[i])
+ else:
+ count[letters2[i]] = i
+
+ return False if count else True