Stealth Interview
  • Features
  • Pricing
  • Blog
  • Login
  • Sign up

Leetcode #2311: Longest Binary Subsequence Less Than or Equal to K

In this guide, we solve Leetcode #2311 Longest Binary Subsequence Less Than or Equal to K in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Greedy, Memoization, String, Dynamic Programming

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

  • Choose a DP state definition.
  • Write the recurrence and base cases.
  • Compute states in the correct order.

Example

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.

Python Solution

class Solution: def longestSubsequence(self, s: str, k: int) -> int: ans = v = 0 for c in s[::-1]: if c == "0": ans += 1 elif ans < 30 and (v | 1 << ans) <= k: v |= 1 << ans ans += 1 return ans

Complexity

The time complexity is O(n)O(n)O(n), where nnn is the length of the string sss. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


Ace your next coding interview

We're here to help you ace your next coding interview.

Subscribe
Stealth Interview
© 2026 Stealth Interview®Stealth Interview is a registered trademark. All rights reserved.
Product
  • Blog
  • Pricing
Company
  • Terms of Service
  • Privacy Policy