Leetcode 567. Permutation in String 双指针

Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.

 

Example 1:

Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").

Example 2:

Input:s1= "ab" s2 = "eidboaoo"
Output: False

 

Note:

1. The input strings only contain lower case letters.
2. The length of both given strings is in range [1, 10,000].

 

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Take care the borderlines:

from collections import Counter

class Solution:
    def checkInclusion(self, s1, s2):
        l1, l2 = len(s1), len(s2)
        if (l1 > l2): #bug1
            return False
        counter = Counter(s1)
        valid_count,pre = 0,0
        dict = {}
        for i in range(l2):
            ch = s2[i]
            dict[ch] = 1 if ch not in dict else dict[ch]+1
            if (ch in counter and dict[ch] <= counter[ch]):
                valid_count += 1
            while (valid_count >= l1):
                if (i-pre+1 == l1 and valid_count == l1): #bug2
                    return True
                to_remove = s2[pre]
                dict[to_remove] -= 1
                if (to_remove in counter and dict[to_remove] < counter[to_remove]):
                    valid_count -= 1
                pre += 1
        return False
s = Solution()
print(s.checkInclusion("a", "ab"))