Check If a String Contains All Binary Codes of Size K (M)

题目

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 indicies 0, 1, 3 and 2 respectively.

Example 2:

Input: s = "00110", k = 2
Output: true

Example 3:

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 4:

Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and doesn‘t exist in the array.

Example 5:

Input: s = "0000000001011100", k = 4
Output: false

Constraints:

• 1 <= s.length <= 5 * 10^5
• s consists of 0‘s and 1‘s only.
• 1 <= k <= 20

题意

判断长度为k的二进制0/1的所有组合是否都是s的子串。

思路

将s中每一个长度为k的子串都加入到set中，最后判断set的容量是否是2^k。

代码实现

Java

class Solution {
    public boolean hasAllCodes(String s, int k) {
        Set<String> set = new HashSet<>();
        for (int i = 0; i + k <= s.length(); i++) {
            set.add(s.substring(i, i + k));
            if (set.size() == (1 << k)) return true;
        }
        return false;
    }
}