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
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