How do you solve Longest Binary Subsequence Less Than or Equal to K on LeetCode?

Longest Binary Subsequence Less Than or Equal to K (LeetCode #2311) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Leading zeros can be included in the subsequence without affecting the value.

What is the brute force solution for Longest Binary Subsequence Less Than or Equal to K?

The brute force approach for Longest Binary Subsequence Less Than or Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the binary string and checking each one to see if it represents a binary number less than or equal to k. This approach is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Binary Subsequence Less Than or Equal to K?

Longest Binary Subsequence Less Than or Equal to K uses the following concepts: String, Dynamic Programming, Greedy, Memoization. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Longest Binary Subsequence Less Than or Equal to K?

Clearly explain your thought process and how you arrived at your approach. Consider edge cases, such as strings with only '0's or '1's. Practice generating subsequences and understanding their binary values.

What are common mistakes when solving Longest Binary Subsequence Less Than or Equal to K?

Not considering leading zeros when counting the length of subsequences. Failing to check if the binary number exceeds k during construction.

#2311

Longest Binary Subsequence Less Than or Equal to K

Medium

StringDynamic ProgrammingGreedyMemoizationGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach focuses on constructing the longest valid subsequence directly by counting zeros and ones, while ensuring the binary number formed is less than or equal to k. This avoids the need to generate all subsequences.

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Algorithm

4 steps

1Step 1: Count the number of '1's in the string s that can be included without exceeding k.
2Step 2: Count the number of '0's in the string s.
3Step 3: If the count of '1's is sufficient to form a number less than or equal to k, include all zeros and those '1's.
4Step 4: If not, include as many '1's as possible while ensuring the binary number remains <= k.

solution.py14 lines

1def longestSubsequence(s, k):
2    count_ones = s.count('1')
3    count_zeros = s.count('0')
4    if count_ones == 0:
5        return count_zeros
6    binary_value = 0
7    length = 0
8    for i in range(len(s)):
9        if s[i] == '1':
10            if binary_value + (1 << (count_ones - 1)) <= k:
11                binary_value += (1 << (count_ones - 1))
12                length += 1
13            count_ones -= 1
14    return length + count_zeros

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Complexity note: The time complexity is O(n) because we only traverse the string a couple of times. The space complexity is O(1) as we only use a few integer variables to keep track of counts and lengths.

1Leading zeros can be included in the subsequence without affecting the value.
2The binary representation grows exponentially, so careful counting of bits is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.