# Leetcode - Check If a String Can Break Another String Solution

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.

## Solution in Python

```
class Solution:
def checkIfCanBreak(self, s1: str, s2: str) -> bool:
s1 = sorted(s1)
s2 = sorted(s2)
return all((x<=y for x,y in zip(s1,s2))) or all((x<=y for x,y in zip(s2,s1)))
```