Introduction

In this blog post, we will explore how to solve the LeetCode problem "1016" 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 a positive integer n, return true if the binary representation of all the integers in the range [1, n] are substrings of s, or false otherwise. A substring is a contiguous sequence of characters within a string. Example 1: Input: s = "0110", n = 3 Output: true Example 2: Input: s = "0110", n = 4 Output: false Constraints: 1 <= s.length <= 1000 s[i] is either '0' or '1'. 1 <= n <= 109

Explanation

Here's the solution:

• Iterate and Check: Loop through numbers from n down to n/2. This optimization significantly reduces the number of iterations because if a larger number's binary representation is a substring, a smaller number's binary representation (especially those smaller than half the larger number) is likely to be a substring as well. By starting from n and going down to n/2, we can quickly identify cases where larger numbers are missing.

• Binary Conversion and Substring Check: Convert each number to its binary representation. Check if this binary string is a substring of the given string s.

• Early Exit: If any number's binary representation is not a substring of s, return False immediately. If all numbers pass the check, return True.

• Complexity: The runtime complexity is O(n * log(n) + s.length). The space complexity is O(log(n)).

Code

    def queryString(s: str, n: int) -> bool:
    """
    Given a binary string s and a positive integer n, return true if the binary representation of all the integers in the range [1, n] are substrings of s, or false otherwise.
    """
    for i in range(n, n // 2, -1):
        binary_i = bin(i)[2:]  # Convert i to binary string, remove "0b" prefix
        if binary_i not in s:
            return False
    return True