leetcode1461. Check If a String Contains All Binary Codes of Size K

Posted 2020-12-10 seyjs

题目如下：

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

解题思路：本题采用逆向思维，先计算出s中长度为k的子串，去重后判断总数是否为2^k即可。

代码如下：

class Solution(object):
    def hasAllCodes(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: bool
        """
        dic = {}
        subs = ‘‘
        for i in range(len(s)):
            if len(subs) < k:
                subs += s[i]
            else:
                subs = subs[1:] + s[i]
            if len(subs) == k:
                dic[subs] = 1
        return len(dic) == 2**k