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closeStrings.py
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closeStrings.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"
# a 2 b 3 c 1
# a 1 b 2 c 3
# 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"
class Solution(object):
def closeStrings(self, str1, str2):
"""
:type word1: str
:type word2: str
:rtype: bool
"""
import collections
counterStr1 = collections.Counter(str1)
counterStr2 = collections.Counter(str2)
keys1 = set(counterStr1.keys())
keys2 = set(counterStr2.keys())
count1 = list(counterStr1.values())
count2 = list(counterStr2.values())
count1.sort()
count2.sort()
if keys1 == keys2 and count1 == count2:
return True
return False
sol = Solution()
res = sol.closeStrings("a", "aa")
print(res)