Skip to content

1657. Determine if Two Strings Are Close

Description

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"

 

Constraints:

  • 1 <= word1.length, word2.length <= 105
  • word1 and word2 contain only lowercase English letters.

 

Solutions

Solution: Greedy

  • Time complexity: O(nlogn)
  • Space complexity: O(n)

 

JavaScript

js
/**
 * @param {string} word1
 * @param {string} word2
 * @return {boolean}
 */
const closeStrings = function (word1, word2) {
  if (word1.length !== word2.length) return false;
  const CODE_BASE = 'a'.charCodeAt(0);
  const word1Count = Array.from({length: 26}).fill(0);
  const word2Count = Array.from({length: 26}).fill(0);

  for (let index = 0; index < word1.length; index++) {
    const charCode1 = word1.charCodeAt(index);
    const charCode2 = word2.charCodeAt(index);

    word1Count[charCode1 - CODE_BASE] += 1;
    word2Count[charCode2 - CODE_BASE] += 1;
  }
  for (let index = 0; index < 26; index++) {
    if (word1Count[index] && !word2Count[index]) return false;
    if (word2Count[index] && !word1Count[index]) return false;
  }

  word1Count.sort((a, b) => b - a);
  word2Count.sort((a, b) => b - a);

  for (let index = 0; index < 26; index++) {
    if (word1Count[index] !== word2Count[index]) return false;
  }
  return true;
};

Released under the MIT license