Check If a String Contains All Binary Codes

Tested Python solution for LeetCode 1461 with 25 pytest cases. Generate a practice environment with lcpy.

LeetCode 1461, Medium. Topics: Hash Table, String, Bit Manipulation, Rolling Hash, Hash Function, Directed Acyclic Graph. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 25 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1461   # by problem number
lcpy gen -s check_if_a_string_contains_all_binary_codes_of_size_k   # by problem name

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Problem

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.

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Examples

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.

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.

Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and does not exist in the array.

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Constraints

• 1 <= s.length <= 5 * 10^5
• s[i] is either '0' or '1'.
• 1 <= k <= 20

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n) each window hashed once; Space: O(2^k) for the set
    def has_all_codes(self, s: str, k: int) -> bool:
        need = 1 << k
        if len(s) < need + k - 1:
            return False
        seen = set()
        for i in range(len(s) - k + 1):
            seen.add(s[i : i + k])
            if len(seen) == need:
                return True
        return False

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Complexity

Time	Space
O(n) each window hashed once; Space: O(2^k) for the set	-

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Tags

NeetCode All.