[leetcode] 1433. Check If a String Can Break Another String

Description

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 or vice-versa.

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.

分析

题目的意思是：给定两个字符串s1和s2，问是否存在一个排列s1中的每个字符都大于s2，相反也可以。我的办法很直接，如果把s1和s2的字母从小到大进行排序，然后一个一个的比较就知道是否是都大于还是都小于了，对于相同字符的位置，跳过去就行了。如果既存在大于的情况又存在小于的情况，直接返回False就可以。

我看了一下其他人的答案，思路差不多，都是先进行排序。

代码

class Solution:
    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
        arr1=list(s1)
        arr2=list(s2)
        arr1.sort()
        arr2.sort()
        n=min(len(arr1),len(arr2))
        cnt1=0
        cnt2=0
        for i in range(n):
            if(arr1[i]<arr2[i]):
                cnt1+=1
            elif(arr1[i]>arr2[i]):
                cnt2+=1
            if(cnt1 and cnt2):
                return False
        return True

参考文献

​​[LeetCode] Python 3 Two Solutions Explained​​