1657. Determine if Two Strings Are Close #

Problem #

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 <= 105
word1 and word2 contain only lowercase English letters.

Problem Summary #

If one string can be obtained from another using the following operations, the two strings are considered close:

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 these two operations on either string as many times as necessary. Given two strings, word1 and word2. If word1 and word2 are close, return true; otherwise, return false.

Solution Ideas #

Determine whether two strings are “close”. The definition of “close” is whether one string can be transformed into another string by swapping two characters or by swapping two letters. If such a transformation exists, they are “close”.
First count the frequencies of the 26 letters in the two strings. If there are frequencies that do not match, return false directly. When the frequencies are the same, sort them in ascending order, then scan again to determine whether the frequencies are the same.
Pay attention to several special cases: when the frequencies are the same, further determine whether the letter swap is valid and exists; if a letter does not exist, output false. For example, case 5 in the test file. The number of frequency occurrences is the same, but the frequencies are different. For example, case 6 in the test file.

Code #

package leetcode

import (
	"sort"
)

func closeStrings(word1 string, word2 string) bool {
	if len(word1) != len(word2) {
		return false
	}
	freqCount1, freqCount2 := make([]int, 26), make([]int, 26)
	for _, c := range word1 {
		freqCount1[c-97]++
	}
	for _, c := range word2 {
		freqCount2[c-97]++
	}
	for i := 0; i < 26; i++ {
		if (freqCount1[i] == freqCount2[i]) ||
			(freqCount1[i] > 0 && freqCount2[i] > 0) {
			continue
		}
		return false
	}
	sort.Ints(freqCount1)
	sort.Ints(freqCount2)
	for i := 0; i < 26; i++ {
		if freqCount1[i] != freqCount2[i] {
			return false
		}
	}
	return true
}