1657. Determine if Two Strings Are Close.md

1657. Determine if Two Strings Are Close

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leetcode Daily Challenge on January 22th, 2021.

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Difficulty : Medium

Related Topics : Greedy

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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 <= 10^5
• word1 and word2 contain only lowercase English letters.

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Solution

• mine
  • Java
    • Runtime: 12 ms, faster than 98.55%, Memory Usage: 40 MB, less than 50.54% of Java online submissions

      public boolean closeStrings(String word1, String word2) {
          if(word1.length() != word2.length()) return false;
          int[] arr1 = new int[26];
          int[] arr2 = new int[26];
          for(char c : word1.toCharArray()){
              arr1[c - 'a']++;
          }
          for(char c : word2.toCharArray()){
              arr2[c - 'a']++;
          }
          int diffCount = 0;
          Map<Integer,Integer> map = new HashMap<>();
          for(int i : arr1){
              if(i != 0){
                  diffCount++;
                  map.put(i, map.getOrDefault(i, 0) + 1);
              }
          }
          for(int i = 0; i < arr2.length; i++){
              if(arr2[i] != 0){
                  if(arr1[i] == 0) return false;
                  diffCount--;
                  int v = map.getOrDefault(arr2[i], 0);
                  if(v == 0) return false;
                  else if(v == 1) map.remove(arr2[i]);
                  else map.put(arr2[i], v - 1);
              }
          }
          return diffCount == 0 && map.size() == 0;
      }
      

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• the most votes

• Runtime: 14 ms, faster than 74.88%, Memory Usage: 39.7 MB, less than 66.88% of Java online submissions

  public boolean closeStrings(String word1, String word2) {
      if(word1.length() != word2.length())
          return false;
  
      int[] f1 = new int[26];
      int[] v1 = new int[26];
      for(char c : word1.toCharArray()) {
          f1[c - 'a']++;
          v1[c - 'a'] = 1;
      }
  
      int[] f2 = new int[26];
      int[] v2 = new int[26];
      for(char c : word2.toCharArray()) {
          f2[c - 'a']++;
          v2[c - 'a'] = 1;
      }
  
      Arrays.sort(f1);
      Arrays.sort(f2);
  
      return Arrays.equals(f1, f2) && Arrays.equals(v1, v2);
  }
  

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