README_EN.md

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 * 105
• s[i] is either '0' or '1'.
• 1 <= k <= 20

Solutions

Python3

class Solution:
    def hasAllCodes(self, s: str, k: int) -> bool:
        counter = 1 << k
        exists = set()
        for i in range(k, len(s) + 1):
            if s[i - k: i] not in exists:
                exists.add(s[i - k: i])
                counter -= 1
            if counter == 0:
                return True
        return False

Java

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

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