Formatted question description: https://leetcode.ca/all/1657.html

# 1657. Determine if Two Strings Are Close (Medium)

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.

Related Topics:
Greedy

Similar Questions:

## Solution 1.

1. Both strings should have the same set of characters.
2. Both strings have the same frequence of characters.
// OJ: https://leetcode.com/problems/determine-if-two-strings-are-close/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
bool closeStrings(string a, string b) {
if (a.size() != b.size()) return false;
array<int, 26> ca = {}, cb = {}, sa = {}, sb = {};
for (char c : a) ca[c - 'a']++, sa[c - 'a'] = 1;
for (char c : b) cb[c - 'a']++, sb[c - 'a'] = 1;
sort(begin(ca), end(ca));
sort(begin(cb), end(cb));
return ca == cb && sa == sb;
}
};


Java

class Solution {
public boolean closeStrings(String word1, String word2) {
if (word1.length() != word2.length())
return false;
int[] counts1 = new int;
int[] counts2 = new int;
int length = word1.length();
for (int i = 0; i < length; i++) {
char c1 = word1.charAt(i), c2 = word2.charAt(i);
counts1[c1 - 'a']++;
counts2[c2 - 'a']++;
}
for (int i = 0; i < 26; i++) {
if (counts1[i] == 0 ^ counts2[i] == 0)
return false;
}
Arrays.sort(counts1);
Arrays.sort(counts2);
return Arrays.equals(counts1, counts2);
}
}