Find Kth Bit in Nth Binary String | Detailed Recursion | Dry Run | Leetcode 1545 | codestorywithMIK
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This is the 9th Video of our Playlist "Recursion : Popular Interview Problems" by codestorywithMIK
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In this video we will try to solve a good Recursion problem : Find Kth Bit in Nth Binary String | Detailed Recursion | Detailed Dry Run | Leetcode 1545 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Find Kth Bit in Nth Binary String | Detailed Recursion | Detailed Dry Run | Leetcode 1545 | codestorywithMIK
Company Tags : will update later
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Recursion/Find Kth Bit in Nth Binary String.cpp
Leetcode Link : https://leetcode.com/problems/find-the-k-th-character-in-string-game-ii/description/
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Summary :
The approach for solving the problem of finding the k-th bit in the n-th binary sequence follows these steps:
Recursive Structure: The binary sequence at level n is built recursively by taking the sequence at level n-1, adding a '1', and then appending the reverse of n-1 where each bit is flipped. This pattern can be leveraged to recursively find the k-th bit.
Base Case: When n = 1, the sequence is just '0', so if the search reaches this level, the result is '0'.
Midpoint Check: The sequence length at level n is 2^n - 1. The midpoint is always '1'. If k is exactly at this midpoint, return '1'.
First Half: If k lies in the first half of the sequence, the answer can be found by recursively searching the corresponding position in the sequence for n-1.
Second Half: If k is in the second half, it corresponds to a mirrored and flipped bit of the first half. Recursively find the corresponding bit in the first half and flip it.
This approach efficiently uses recursion to handle the reverse and flip operations without generating the entire sequence explicitly.
✨ Timelines✨
00:00 - Introduction
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