[leetcode] 1461. Check If a String Contains All Binary Codes of Size K
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Description
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
分析
题目的意思是：给定一个字符串和k，看字符串中的所有为k的子串都是s的子串。我的思路也是很简单，直接截取s中为k的子串存放到字典中，最后判断字典的长度会不会时2**k就行了。
代码
class Solution:
def hasAllCodes(self, s: str, k: int) -> bool:
n=len(s)
d=collections.defaultdict(int)
for i in range(n-k+1):
d[s[i:i+k]]+=1
# print(d)
return len(d)==(2**k)
参考文献
[LeetCode] soluion
