Leetcode 567 Permutation in String

Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.

In other words, return true if one of s1’s permutations is the substring of s2.

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

Input: s1 = "ab", s2 = "eidboaoo"
Output: false

• For each window representing a substring of s2 of length len(s1), we want to check if the count of the window is equal to the count of s1. Here, the count of a string is the list of: [the number of a’s it has, the number of b’s,… , the number of z’s.]

We can maintain the window by deleting the value of s2[i - len(s1)] when it gets larger than len(s1). After, we only need to check if it is equal to the target. Working with list values of [0, 1,…, 25] instead of ‘a’-‘z’ makes it easier to count later.

class Solution:
    def checkInclusion(self, s1, s2):
        A = [ord(x) - ord('a') for x in s1]
        B = [ord(x) - ord('a') for x in s2]
    
        target = [0] * 26
        for x in A:
            target[x] += 1
    
        window = [0] * 26
        for i, x in enumerate(B):
            window[x] += 1
            if i >= len(A):
                window[B[i - len(A)]] -= 1
            if window == target:
                return True
        return False