1433. Check If a String Can Break Another String

Problem

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 (Java)

class Solution {
    public boolean checkIfCanBreak(String s1, String s2) {
        if (s1 == null && s2 == null || Objects.requireNonNull(s1).length() == 1) {
            return true;
        }
        int[] count1 = new int[26];
        int[] count2 = new int[26];

        for (int i = s1.length() - 1; i >= 0; i--) {
            count1[s1.charAt(i) - 'a']++;
            count2[s2.charAt(i) - 'a']++;
        }
        boolean isS1Greater = count1[25] >= count2[25];
        boolean isS2Greater = count2[25] >= count1[25];

        for (int i = 24; (isS1Greater || isS2Greater) && i >= 0; i--) {
            count1[i] += count1[i + 1];
            count2[i] += count2[i + 1];

            isS1Greater = isS1Greater && count1[i] >= count2[i];
            isS2Greater = isS2Greater && count2[i] >= count1[i];
        }
        return isS1Greater || isS2Greater;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).