1657.Determine if Two Strings Are Close

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

1657. 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, 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.

範例

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.

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

Kobe 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)

玉山 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。
Marsgoat Dec 2, 2022

Reference

回到題目列表