1461-check-if-a-string-contains-all-binary-codes-of-size-k

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Description

Given a binary string s and an integer k, return true if every binary code of length k is a substring of s. Otherwise, return false.

 

Example 1:

Input: s = "00110110", k = 2
Output: true
Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indices 0, 1, 3 and 2 respectively.

Example 2:

Input: s = "0110", k = 1
Output: true
Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring. 

Example 3:

Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and does not exist in the array.

 

Constraints:

• 1 <= s.length <= 5 * 105
• s[i] is either '0' or '1'.
• 1 <= k <= 20

Solution(Python)

class Solution:
    def hasAllCodes(self, s: str, k: int) -> bool:
        return self.optimal(s, k)

    # Time Complexity: O(n*k)
    # Space Complexity: O(n*k)
    def bruteforce(self, s: str, k: int) -> bool:
        subsets = {s[i: i + k] for i in range(len(s) - k + 1)}
        return len(subsets) == 1 << k

    # Time Complexity: O(n)
    # Space Complexity: O(2^k)
    def optimal(self, s: str, k: int) -> bool:
        need = 1 << k
        got = [False] * need
        all_one = need - 1
        hash_val = 0

        for i in range(len(s)):
            hash_val = ((hash_val << 1) & all_one) | int(s[i])
            if i >= k - 1 and got[hash_val] is False:
                got[hash_val] = True
                need -= 1
                if need == 0:
                    return True
        return False