LeetCode #2311 — MEDIUM

Longest Binary Subsequence Less Than or Equal to K

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Greedy

Example 1

"1001010"
5

Example 2

"00101001"
1

Core Insight

What unlocks the optimal approach

Choosing a subsequence from the string is equivalent to deleting all the other digits.
If you were to remove a digit, which one should you remove to reduce the value of the string?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2311: Longest Binary Subsequence Less Than or Equal to K
class Solution {
    public int longestSubsequence(String s, int k) {
        int ans = 0, v = 0;
        for (int i = s.length() - 1; i >= 0; --i) {
            if (s.charAt(i) == '0') {
                ++ans;
            } else if (ans < 30 && (v | 1 << ans) <= k) {
                v |= 1 << ans;
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2311: Longest Binary Subsequence Less Than or Equal to K
func longestSubsequence(s string, k int) (ans int) {
	for i, v := len(s)-1, 0; i >= 0; i-- {
		if s[i] == '0' {
			ans++
		} else if ans < 30 && (v|1<<ans) <= k {
			v |= 1 << ans
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2311: Longest Binary Subsequence Less Than or Equal to K
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

// Accepted solution for LeetCode #2311: Longest Binary Subsequence Less Than or Equal to K
impl Solution {
    pub fn longest_subsequence(s: String, k: i32) -> i32 {
        let mut ans = 0;
        let mut v = 0;
        let s = s.as_bytes();
        for i in (0..s.len()).rev() {
            if s[i] == b'0' {
                ans += 1;
            } else if ans < 30 && (v | (1 << ans)) <= k {
                v |= 1 << ans;
                ans += 1;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2311: Longest Binary Subsequence Less Than or Equal to K
function longestSubsequence(s: string, k: number): number {
    let ans = 0;
    for (let i = s.length - 1, v = 0; ~i; --i) {
        if (s[i] == '0') {
            ++ans;
        } else if (ans < 30 && (v | (1 << ans)) <= k) {
            v |= 1 << ans;
            ++ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.