2839. Check if Strings Can be Made Equal With Operations I

/**
 * Problem: LeetCode 2839 - Check if Strings Can be Made Equal With Operations I
 * Difficulty: Easy
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public boolean canBeEqual(String s1, String s2) {
        boolean evenMatch = (s1.charAt(0) == s2.charAt(0) && s1.charAt(2) == s2.charAt(2)) ||
                            (s1.charAt(0) == s2.charAt(2) && s1.charAt(2) == s2.charAt(0));

        boolean oddMatch = (s1.charAt(1) == s2.charAt(1) && s1.charAt(3) == s2.charAt(3)) ||
                           (s1.charAt(1) == s2.charAt(3) && s1.charAt(3) == s2.charAt(1));

        return evenMatch && oddMatch;
    }
}

"""
Problem: LeetCode 2839 - Check if Strings Can be Made Equal With Operations I
Difficulty: Easy
Language: Python
Author: CodeByTushu
"""
class Solution:
    def canBeEqual(self, s1: str, s2: str) -> bool:
        even_match = (
            (s1[0] == s2[0] and s1[2] == s2[2]) or
            (s1[0] == s2[2] and s1[2] == s2[0])
        )
        odd_match = (
            (s1[1] == s2[1] and s1[3] == s2[3]) or
            (s1[1] == s2[3] and s1[3] == s2[1])
        )
        return even_match and odd_match

/**
 * Problem: LeetCode 2839 - Check if Strings Can be Made Equal With Operations I
 * Difficulty: Easy
 * Language: C++
 * Author: CodeByTushu
 */
#include <string>

using namespace std;

class Solution {
public:
    bool canBeEqual(string s1, string s2) {
        bool evenMatch = (s1[0] == s2[0] && s1[2] == s2[2]) ||
                         (s1[0] == s2[2] && s1[2] == s2[0]);

        bool oddMatch = (s1[1] == s2[1] && s1[3] == s2[3]) ||
                        (s1[1] == s2[3] && s1[3] == s2[1]);

        return evenMatch && oddMatch;
    }
};

/**
 * Problem: LeetCode 2839 - Check if Strings Can be Made Equal With Operations I
 * Difficulty: Easy
 * Language: JavaScript
 * Author: CodeByTushu
 */
var canBeEqual = function(s1, s2) {
    const evenMatch = (s1[0] === s2[0] && s1[2] === s2[2]) ||
                      (s1[0] === s2[2] && s1[2] === s2[0]);

    const oddMatch = (s1[1] === s2[1] && s1[3] === s2[3]) ||
                     (s1[1] === s2[3] && s1[3] === s2[1]);

    return evenMatch && oddMatch;
};