567-permutation-in-string¶

Description¶

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.

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

Constraints:

1 <= s1.length, s2.length <= 10⁴
s1 and s2 consist of lowercase English letters.

Solution(Python)¶

class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        return self.slidingWindow(s1, s2)

    """
    Time Complexity: O(n!)
    Space Complexity: O(n^2)
    """

    def bruteforce(self, s1: str, s2: str) -> bool:
        self.res = False

        def permute(s1, s2, l):
            if l == len(s1):
                if "".join(s1) in "".join(s2):
                    self.res = True
            else:
                for i in range(l, len(s1)):
                    s1[l], s1[i] = s1[i], s1[l]
                    permute(s1, s2, l + 1)
                    s1[l], s1[i] = s1[i], s1[l]

        permute(list(s1), list(s2), 0)
        return self.res

    """
    we can use either hashmap or an array of size 26 to store the frequency of array
    Time Complexity: O(m*(n-m))
    Space Complexity: O(1)
    """

    def extraDataStructure(self, s1: str, s2: str) -> bool:
        s1Freq = [0] * 26

        for ch in s1:
            s1Freq[ord(ch) - ord("a")] += 1

        n = len(s2)
        m = len(s1)

        for i in range(n - m + 1):
            s2Freq = [0] * 26
            for j in range(m):
                ch = s2[i + j]
                s2Freq[ord(ch) - ord("a")] += 1
            if s1Freq == s2Freq:
                return True
        return False

    """
    Time Complexity: O(m+(n-m))
    Space Complexity: O(1)
    """

    def slidingWindow(self, s1: str, s2: str) -> bool:
        if len(s1) > len(s2):
            return False

        s1Freq = [0] * 26
        s2Freq = [0] * 26

        for i in range(len(s1)):
            s1Freq[ord(s1[i]) - ord("a")] += 1
            s2Freq[ord(s2[i]) - ord("a")] += 1

        count = 0

        for i in range(26):
            if s1Freq[i] == s2Freq[i]:
                count += 1

        for i in range(len(s2) - len(s1)):
            r = ord(s2[i + len(s1)]) - ord("a")
            l = ord(s2[i]) - ord("a")
            if count == 26:
                return True
            s2Freq[r] += 1
            if s2Freq[r] == s1Freq[r]:
                count += 1
            elif s2Freq[r] == s1Freq[r] + 1:
                count -= 1
            s2Freq[l] -= 1
            if s2Freq[l] == s1Freq[l]:
                count += 1
            elif s2Freq[l] == s1Freq[l] - 1:
                count -= 1

        return count == 26