1657.Determine if Two Strings Are Close
===

###### tags: `Medium`,`String`,`Hash Table`,`Sorting`

[1657. Determine if Two Strings Are Close](https://leetcode.com/problems/determine-if-two-strings-are-close/)

### 題目描述

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, a++b++cd++e++ -> a++e++cd++b++
* Operation 2: Transform **every** occurrence of one **existing** character into another **existing** character, and do the same with the other character.
* > For example, ++aa++c++abb++ -> ++bb++c++baa++ (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*.

### 範例

**Example 1:**

Input: word1 = "abc", word2 = "bca"
Output: true
Explanation: You can attain word2 from word1 in 2 operations.

`Apply Operation 1:` "a++bc++" -> "a++cb++"
`Apply Operation 1:` "++a++c++b++" -> "++b++c++a++"

**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:` "ca++b++bb++a++" -> "ca++a++bb++b++"
`Apply Operation 2:` "++c++aa++bbb++" -> "++b++aa++ccc++"
`Apply Operation 2:` "++baa++ccc" -> "++abb++ccc"

**Constraints**:

* 1 <= `word1.length`, `word2.length`<= 10^5^
* `word1` and `word2` contain only lowercase English letters.

### 解答

#### Python

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if len(word1) != len(word2): return False
        
        dict1, dict2 = {}, {}
        
        for word in word1:
            dict1[word] = dict1.get(word, 0) + 1
            
        for word in word2:
            dict2[word] = dict2.get(word, 0) + 1
            
        val1, val2 = sorted(dict1.values()), sorted(dict2.values())
        
        return set(word1) == set(word2) and val1 == val2

> [name=Kobe] [time= Dec 2, 2022]

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        def stat(word):
            Q = dict()
            for x in word:
                if x in Q:
                    Q[x] += 1
                else:
                    Q[x] = 1
            
            amount = [ v for i,(k,v) in enumerate(Q.items())]
            word_set = [ k for i,(k,v) in enumerate(Q.items())]
            amount = sorted(amount)
            word_set = sorted(word_set)
            #print(word_set)
            #print(amount)
            return amount+word_set
        
        return stat(word1) == stat(word2)

> [name=玉山] [time= Dec 2, 2022]

#### Javascript

function closeStrings(word1, word2) {
  if (word1.length !== word2.length) return false;
  const map1 = new Map();
  const map2 = new Map();
  for (let i = 0; i < word1.length; i++) {
    map1.set(word1[i], (map1.get(word1[i]) || 0) + 1);
    map2.set(word2[i], (map2.get(word2[i]) || 0) + 1);
  }
    
  // 要有相同的字母
  if (map1.size !== map2.size) return false;
  for (const key of map1.keys()) {
    if (!map2.has(key)) return false;
  }

  // 且字母個數分布要一樣
  const arr1 = Array.from(map1.values()).sort((a, b) => a - b);
  const arr2 = Array.from(map2.values()).sort((a, b) => a - b);
  for (let i = 0; i < arr1.length; i++) {
    if (arr1[i] !== arr2[i]) return false;
  }

  return true;
}

> 順哥：就char set要一樣，個數分布要一樣。
> 感謝順哥給的提示，感恩順哥讚嘆順哥。
> [name=Marsgoat] [time= Dec 2, 2022]

### Reference

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