Two strings are considered close if you can attain one from the other using the following operations:
abcde -> aecdb
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 {
public:
bool closeStrings(string word1, string word2) {
unordered_map map1,map2;
if(word1.length()!=word2.length()) return false;
for(int i=0; i vect1,vect2;
for(auto i:map1) vect1.push_back(i.second);
for(auto i:map2) vect2.push_back(i.second);
sort(vect1.begin(),vect1.end());
sort(vect2.begin(),vect2.end());
return vect1==vect2;
}
};
The purpose of this function is to determine whether two strings, ‘word1’ and ‘word2’, are close under certain conditions. Two words are considered close if they have the same set of characters and the frequencies of these characters are also the same.
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