Introduction

In this blog post, we will explore how to solve the LeetCode problem "2311" 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

You are given a binary string s and a positive integer k. Return the length of the longest subsequence of s that makes up a binary number less than or equal to k. Note: The subsequence can contain leading zeroes. The empty string is considered to be equal to 0. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. Example 1: Input: s = "1001010", k = 5 Output: 5 Explanation: The longest subsequence of s that makes up a binary number less than or equal to 5 is "00010", as this number is equal to 2 in decimal. Note that "00100" and "00101" are also possible, which are equal to 4 and 5 in decimal, respectively. The length of this subsequence is 5, so 5 is returned. Example 2: Input: s = "00101001", k = 1 Output: 6 Explanation: "000001" is the longest subsequence of s that makes up a binary number less than or equal to 1, as this number is equal to 1 in decimal. The length of this subsequence is 6, so 6 is returned. Constraints: 1 <= s.length <= 1000 s[i] is either '0' or '1'. 1 <= k <= 109

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Prioritize including all '0's since they don't increase the numerical value of the subsequence.
  • Iterate through the '1's in reverse order. For each '1', check if including it in the current subsequence would exceed k. If not, include it.
  • The length of the final subsequence is the sum of the '0's and the included '1's.

• Complexity:

  • Runtime: O(n), where n is the length of the string s.
  • Storage: O(1)

Code

    def longestSubsequence(s: str, k: int) -> int:
    """
    Finds the length of the longest subsequence of s that makes up a binary
    number less than or equal to k.
    """
    zeros = s.count('0')
    ones = 0
    num = 0
    power = 1
    for i in range(len(s) - 1, -1, -1):
        if s[i] == '1':
            if power <= k - num:
                num += power
                ones += 1
            if power > k:
              break
            power *= 2

    return zeros + ones