LeetCode: Check If a String Can Break Another String

Identity number which appears exactly once.

Similar Problems:

• CheatSheet: LeetCode For Code Interview
• CheatSheet: Common Code Problems & Follow-ups
• Tag: #greedy, #string, #twopointer

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:

## https://code.dennyzhang.com/check-if-a-string-can-break-another-string
## Basic Ideas: two pointer + greedy
##
## The order doesn't matter.
## For the same characters in both two string, remove them
##
## Complexity: Time ?, Space ?
class Solution:
    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
        def mycheck(s1, s2):
            n = len(s1)
            cnt1, cnt2 = [0]*26, [0]*26
            for ch in s1:
                cnt1[ord(ch)-ord('a')] += 1
            for ch in s2:
                index = ord(ch)-ord('a')
                if cnt1[index]>0:
                    cnt1[index] -= 1
                else:
                    cnt2[index] += 1
            l1, l2 = [], []
            for i in range(26):
                if cnt1[i] != 0:
                    l1.extend([chr(ord('a')+i)]*cnt1[i])
                if cnt2[i] != 0:
                    l2.extend([chr(ord('a')+i)]*cnt2[i])
            for i in range(len(l1)):
                if l1[i]<l2[i]: return False
            return True
        return mycheck(s1, s2) or mycheck(s2, s1)