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LeetCode 2311: Longest Binary Subsequence Less Than or Equal to K | Java Solution

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🚀 LeetCode 2311 - Longest Binary Subsequence Less Than or Equal to K | Java Solution Explained
In this video, we solve LeetCode problem 2311 titled Longest Binary Subsequence Less Than or Equal to K using a greedy approach combined with bit manipulation in Java.
🧠 Problem Summary:
You are given a binary string s and an integer k. The goal is to find the maximum length of a subsequence of s such that the binary value of the selected subsequence is less than or equal to k.
🔍 What We Cover:
Traverse the string from right to left to prioritise lower power bits.
Always include '0' characters since they don't affect the sum.
Include '1' characters only if their addition does not exceed the value of k.
Use bit shifting to efficiently compute binary values.
Stop processing further bits once the total exceeds k.
💡 Key Concepts:
Greedy strategy from the least significant bit
Bit manipulation with bit shifting (1L shifted left by index)
Subsequence selection to maximise length under constraint
🧾 Time Complexity: O(n)
🧾 Space Complexity: O(1)
This is a great problem to practice greedy thinking with binary numbers and understand how to efficiently handle constraints in subsequence problems.
👉 Don’t forget to like, comment, and subscribe for more Java-based coding tutorials!
#leetcode #greedy #bitmanipulation #javaprogramming #subsequence #codingxkrishna
In this video, we solve LeetCode problem 2311 titled Longest Binary Subsequence Less Than or Equal to K using a greedy approach combined with bit manipulation in Java.
🧠 Problem Summary:
You are given a binary string s and an integer k. The goal is to find the maximum length of a subsequence of s such that the binary value of the selected subsequence is less than or equal to k.
🔍 What We Cover:
Traverse the string from right to left to prioritise lower power bits.
Always include '0' characters since they don't affect the sum.
Include '1' characters only if their addition does not exceed the value of k.
Use bit shifting to efficiently compute binary values.
Stop processing further bits once the total exceeds k.
💡 Key Concepts:
Greedy strategy from the least significant bit
Bit manipulation with bit shifting (1L shifted left by index)
Subsequence selection to maximise length under constraint
🧾 Time Complexity: O(n)
🧾 Space Complexity: O(1)
This is a great problem to practice greedy thinking with binary numbers and understand how to efficiently handle constraints in subsequence problems.
👉 Don’t forget to like, comment, and subscribe for more Java-based coding tutorials!
#leetcode #greedy #bitmanipulation #javaprogramming #subsequence #codingxkrishna
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