README.md

1433. 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

Constraints:

• s1.length == n
• s2.length == n
• 1 <= n <= 10^5
• All strings consist of lowercase English letters.

Solutions (Rust)

1. Sort

impl Solution {
    pub fn check_if_can_break(s1: String, s2: String) -> bool {
        let mut s1 = s1.into_bytes();
        let mut s2 = s2.into_bytes();
        s1.sort_unstable();
        s2.sort_unstable();
        let s1s2 = s1.iter().zip(s2.iter()).collect::<Vec<_>>();

        s1s2.iter().all(|tup| tup.0 >= tup.1) || s1s2.iter().all(|tup| tup.0 <= tup.1)
    }
}