Formatted question description: https://leetcode.ca/all/1433.html
1433. Check If a String Can Break Another String
Level
Medium
Description
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.
Solution
Convert s1
and s2
into arrays of type char
, which are array1
and array2
respectively, and sort array1
and array2
. Then compare each pair of characters in array1
and array2
. If there exists two indices i
and j
such that array1[i] < array2[i]
and array1[j] > array2[j]
, then return false
. Otherwise, return true
.
class Solution {
public boolean checkIfCanBreak(String s1, String s2) {
char[] array1 = s1.toCharArray();
char[] array2 = s2.toCharArray();
Arrays.sort(array1);
Arrays.sort(array2);
int difference = 0;
int length = array1.length;
for (int i = 0; i < length; i++) {
int curDifference = array1[i] - array2[i];
if (curDifference != 0) {
curDifference /= Math.abs(curDifference);
if (curDifference * difference < 0)
return false;
else
difference = curDifference;
}
}
return true;
}
}