LeetCode – Strings – Check If a String Can Break Another String

Given two strings: s1 and s2 with the same size, check if some permutation of string s1 can break some permutation of string s2 or vice-versa (in other words s2 can break s1).

A string x can break string y (both of size n) if x[i] >= y[i] (in alphabetical order) for all i between 0 and n-1.

Example 1:

Input: s1 = "abc", s2 = "xya"
Output: true
Explanation: "ayx" is a permutation of s2="xya" which can break to string "abc" which is a permutation of s1="abc".

Example 2:

Input: s1 = "abe", s2 = "acd"
Output: false 
Explanation: All permutations for s1="abe" are: "abe", "aeb", "bae", "bea", "eab" and "eba" and all permutation for s2="acd" are: "acd", "adc", "cad", "cda", "dac" and "dca". However, there is not any permutation from s1 which can break some permutation from s2 and vice-versa.

Example 3:

Input: s1 = "leetcodee", s2 = "interview"
Output: true

class Solution {
    public boolean checkIfCanBreak(String s1, String s2) {
        if(s1.equals(s2)) return true;
        char[] s1_arr = s1.toCharArray();
        char[] s2_arr = s2.toCharArray();
        
        Arrays.sort(s1_arr);
        Arrays.sort(s2_arr);
        
        boolean result1 = true;
        boolean result2 = true;
        
        for(int i=0;i<s1_arr.length;i++){
            if(s1_arr[i]<s2_arr[i]){
                result1=false;
                break;
            }
        } 
        
        for(int i=0;i<s2_arr.length;i++){
            if(s2_arr[i]<s1_arr[i]){
                result2=false;
                break;
            }
        }
        
        return (result1 || result2);
    }
}