Given a binary string s and an integer k, return trueif every binary code of lengthkis a substring ofs. 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
Solutions
Solution 1: Hash Table
First, for a string $s$ of length $n$, the number of substrings of length $k$ is $n - k + 1$. If $n - k + 1 < 2^k$, then there must exist a binary string of length $k$ that is not a substring of $s$, so we return false.
Next, we traverse the string $s$ and store all substrings of length $k$ in a set $ss$. Finally, we check if the size of the set $ss$ is equal to $2^k$.
The time complexity is $O(n \times k)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.
In Solution 1, we stored all distinct substrings of length $k$, and processing each substring requires $O(k)$ time. We can instead use a sliding window, where each time we add the latest character, we remove the leftmost character from the window. During this process, we use an integer $x$ to store the substring.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.