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
Constraints:
s1.length == n
s2.length == n
1 <= n <= 10^5
All strings consist of lowercase English letters.
class Solution:
    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
        from collections import Counter
        def check_breakable(s1, s2):
          s1_counter = Counter(s1)
          s2_counter = Counter(s2)
          cnt = 0
          for char in 'abcdefghijklmnopqrstuvwxyz':
            cnt += s1_counter[char] - s2_counter[char]
            if cnt < 0:
              return False
          return True
        return check_breakable(s1, s2) == True or check_breakable(s2,s1) == True
class Solution:
    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
        from collections import Counter
        def check_breakable(s1, s2):
          s1_counter = Counter(s1)
          s2_counter = Counter(s2)
          cnt = 0
          for char in 'abcdefghijklmnopqrstuvwxyz':
            cnt += s1_counter[char] - s2_counter[char]
            if cnt < 0:
              return False
          return True
        return check_breakable(s1, s2) == True or check_breakable(s2,s1) == True