Check If a String Contains All Binary Codes of Size K

#leetcode #dsa #theabbie

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:

class Solution:
    def hasAllCodes(self, s: str, k: int) -> bool:
        n = len(s)
        codes = set()
        curr = int(s[0:k], 2)
        codes.add(curr)
        for i in range(n - k):
            curr = (curr << 1) - int(s[i]) * (1 << k) + int(s[i + k])
            codes.add(curr)
        return len(codes) == 1 << k