Introduction

In this blog post, we will explore how to solve the LeetCode problem "1461" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 indices 0, 1, 3 and 2 respectively. Example 2: 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 3: Input: s = "0110", k = 2 Output: false Explanation: The binary code "00" is of length 2 and does not exist in the array. Constraints: 1 <= s.length <= 5 * 105 s[i] is either '0' or '1'. 1 <= k <= 20

Explanation

Here's a breakdown of the solution:

• Key Idea: Iterate through the string s and extract all possible substrings of length k. Convert these substrings into their integer representation. Use a set to efficiently keep track of the unique integer representations encountered.
• Completeness Check: After iterating through s, check if the number of unique integer representations in the set is equal to 2^k. If they are equal, then all binary codes of length k are present as substrings.

• Efficiency: Using a set provides efficient lookups and ensures no duplicate substrings are counted. Integer conversion allows quick comparisons.

• Complexity: O(n) runtime, O(2^k) storage where n is the length of s.

Code

    def hasAllCodes(s: str, k: int) -> bool:
    """
    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.
    """
    seen = set()
    for i in range(len(s) - k + 1):
        sub = s[i:i+k]
        seen.add(int(sub, 2))
    return len(seen) == 2**k