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.
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]
)
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;
}
};
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;
}