No1433.check-if-a-string-can-break-another-string.cs

/*
 * Difficulty:
 * Medium
 *
 * Desc:
 * 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.
*/

public class Solution {
    public bool CheckIfCanBreak(string s1, string s2) {
        char[] arr1 = s1.ToCharArray();
        Array.Sort(arr1);

        char[] arr2 = s2.ToCharArray();
        Array.Sort(arr2);

        int greater = 0;
        for (int i = 0; i < arr1.Length; i += 1) {
            if (arr1[i] == arr2[i]) continue;
            if (arr1[i] > arr2[i] && greater < 0) return false;
            if (arr1[i] < arr2[i] && greater > 0) return false;
            greater = arr1[i] - arr2[i];
        }
        return true;
    }
}