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2839. Check if Strings Can be Made Equal With Operations I
Description
You are given two strings s1
and s2
, both of length 4
, consisting of lowercase English letters.
You can apply the following operation on any of the two strings any number of times:
- Choose any two indices
i
andj
such thatj - i = 2
, then swap the two characters at those indices in the string.
Return true
if you can make the strings s1
and s2
equal, and false
otherwise.
Example 1:
Input: s1 = "abcd", s2 = "cdab" Output: true Explanation: We can do the following operations on s1: - Choose the indices i = 0, j = 2. The resulting string is s1 = "cbad". - Choose the indices i = 1, j = 3. The resulting string is s1 = "cdab" = s2.
Example 2:
Input: s1 = "abcd", s2 = "dacb" Output: false Explanation: It is not possible to make the two strings equal.
Constraints:
s1.length == s2.length == 4
s1
ands2
consist only of lowercase English letters.
Solutions
-
class Solution { public boolean canBeEqual(String s1, String s2) { int[][] cnt = new int[2][26]; for (int i = 0; i < s1.length(); ++i) { ++cnt[i & 1][s1.charAt(i) - 'a']; --cnt[i & 1][s2.charAt(i) - 'a']; } for (int i = 0; i < 26; ++i) { if (cnt[0][i] != 0 || cnt[1][i] != 0) { return false; } } return true; } }
-
class Solution { public: bool canBeEqual(string s1, string s2) { vector<vector<int>> cnt(2, vector<int>(26, 0)); for (int i = 0; i < s1.size(); ++i) { ++cnt[i & 1][s1[i] - 'a']; --cnt[i & 1][s2[i] - 'a']; } for (int i = 0; i < 26; ++i) { if (cnt[0][i] || cnt[1][i]) { return false; } } return true; } };
-
class Solution: def canBeEqual(self, s1: str, s2: str) -> bool: return sorted(s1[::2]) == sorted(s2[::2]) and sorted(s1[1::2]) == sorted( s2[1::2] )
-
func canBeEqual(s1 string, s2 string) bool { cnt := [2][26]int{} for i := 0; i < len(s1); i++ { cnt[i&1][s1[i]-'a']++ cnt[i&1][s2[i]-'a']-- } for i := 0; i < 26; i++ { if cnt[0][i] != 0 || cnt[1][i] != 0 { return false } } return true }
-
function canBeEqual(s1: string, s2: string): boolean { const cnt: number[][] = Array.from({ length: 2 }, () => Array.from({ length: 26 }, () => 0)); for (let i = 0; i < s1.length; ++i) { ++cnt[i & 1][s1.charCodeAt(i) - 97]; --cnt[i & 1][s2.charCodeAt(i) - 97]; } for (let i = 0; i < 26; ++i) { if (cnt[0][i] || cnt[1][i]) { return false; } } return true; }