Two strings are considered close if you can attain one from the other using the following operations:
- Operation 1: Swap any two existing characters.
<ul> <li>For example, <code>a<u>b</u>cd<u>e</u> -> a<u>e</u>cd<u>b</u></code></li> </ul> </li> <li>Operation 2: Transform <strong>every</strong> occurrence of one <strong>existing</strong> character into another <strong>existing</strong> character, and do the same with the other character. <ul> <li>For example, <code><u>aa</u>c<u>abb</u> -> <u>bb</u>c<u>baa</u></code> (all <code>a</code>'s turn into <code>b</code>'s, and all <code>b</code>'s turn into <code>a</code>'s)</li> </ul> </li>
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
andword2
contain only lowercase English letters.
Related Topics:
Greedy
Similar Questions:
- Buddy Strings (Easy)
- Minimum Swaps to Make Strings Equal (Medium)
- Minimum Number of Steps to Make Two Strings Anagram (Medium)
- Both strings should have the same set of characters.
- Both strings have the same frequence of characters.
// OJ: https://leetcode.com/problems/determine-if-two-strings-are-close/
// Author: github.com/lzl124631x
// 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;
}
};