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1657. Determine if Two Strings Are Close

中文文档

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

Python3

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        cnt1, cnt2 = Counter(word1), Counter(word2)
        return sorted(cnt1.values()) == sorted(cnt2.values()) and set(cnt1.keys()) == set(cnt2.keys())

Java

class Solution {
    public boolean closeStrings(String word1, String word2) {
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < word1.length(); ++i) {
            ++cnt1[word1.charAt(i) - 'a'];
        }
        for (int i = 0; i < word2.length(); ++i) {
            ++cnt2[word2.charAt(i) - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if ((cnt1[i] > 0 && cnt2[i] == 0) || (cnt2[i] > 0 && cnt1[i] == 0)) {
                return false;
            }
        }
        Arrays.sort(cnt1);
        Arrays.sort(cnt2);
        for (int i = 0; i < 26; ++i) {
            if (cnt1[i] != cnt2[i]) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    bool closeStrings(string word1, string word2) {
        int cnt1[26]{};
        int cnt2[26]{};
        for (char& c : word1) {
            ++cnt1[c - 'a'];
        }
        for (char& c : word2) {
            ++cnt2[c - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if ((cnt1[i] > 0 && cnt2[i] == 0) || (cnt1[i] == 0 && cnt2[i] > 0)) {
                return false;
            }
        }
        sort(cnt1, cnt1 + 26);
        sort(cnt2, cnt2 + 26);
        for (int i = 0; i < 26; ++i) {
            if (cnt1[i] != cnt2[i]) {
                return false;
            }
        }
        return true;
    }
};

Go

func closeStrings(word1 string, word2 string) bool {
	cnt1 := make([]int, 26)
	cnt2 := make([]int, 26)
	for _, c := range word1 {
		cnt1[c-'a']++
	}
	for _, c := range word2 {
		cnt2[c-'a']++
	}
	for i, v := range cnt1 {
		if (v > 0 && cnt2[i] == 0) || (v == 0 && cnt2[i] > 0) {
			return false
		}
	}
	sort.Ints(cnt1)
	sort.Ints(cnt2)
	for i, v := range cnt1 {
		if v != cnt2[i] {
			return false
		}
	}
	return true
}

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