LeetCode in Kotlin

1433. Check If a String Can Break Another String

Medium

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

class Solution {
    fun checkIfCanBreak(s1: String, s2: String): Boolean {
        if (s1.length == 1) {
            return true
        }
        val count1 = IntArray(26)
        val count2 = IntArray(26)
        for (i in s1.length - 1 downTo 0) {
            count1[s1[i].code - 'a'.code]++
            count2[s2[i].code - 'a'.code]++
        }
        var isS1Greater = count1[25] >= count2[25]
        var isS2Greater = count2[25] >= count1[25]
        var i = 24
        while ((isS1Greater || isS2Greater) && i >= 0) {
            count1[i] += count1[i + 1]
            count2[i] += count2[i + 1]
            isS1Greater = isS1Greater && count1[i] >= count2[i]
            isS2Greater = isS2Greater && count2[i] >= count1[i]
            i--
        }
        return isS1Greater || isS2Greater
    }
}

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