Number of 1 Bits | 4 Approaches | Time Complexity | Apple | Microsoft | Amazon | Leetcode 191
137Whatsapp Community Link : https://www.whatsapp.com/channel/0029Va6kVSjICVfiVdsHgi1A
iPad PDF Notes - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/iPad%20PDF%20Notes/Leetcode-191-Number%20of%201%20Bits.pdf
This is the 7th Video of our Bit Manipulation Playlist.
In this video we will try to solve an easy BUT a very informative Bit manipulation problem - Number of 1 Bits (Leetcode -191).
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 : Number of 1 Bits
Company Tags : Amazon, Apple, Microsoft
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Bit_Magic/Number%20of%201%20Bits.cpp
Leetcode Link : https://leetcode.com/problems/number-of-1-bits/
My DP Concepts Playlist : https://youtu.be/7eLMOE1jnls
My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Interview_DS_Algo
Subscribe to my channel : https://www.youtube.com/@codestorywithMIK
Instagram : https://www.instagram.com/codestorywithmik/
Facebook : https://www.facebook.com/people/codestorywithmik/100090524295846/
Twitter : https://twitter.com/CSwithMIK
Approach-1 Summary :
We are checking whether the i-th bit of the input number n is set to 1 using bitwise operations. n right shift i shifts the bits of n to the right by i positions, bringing the i-th bit to the least significant position. & 1 masks all bits except the least significant one, effectively checking whether the i-th bit is 1. If the i-th bit is 1, increments the count variable.
Approach-2 Summary :
Initiates a while loop that continues until n becomes zero. Within the loop, the expression n = (n & (n - 1)) is used to turn off the rightmost set bit in n. This operation effectively removes one set bit in each iteration. The count variable is incremented for each set bit removed.
Approach-3 Summary :
Initiates a while loop that continues until n becomes zero.
Within the loop, count is incremented by the result of the modulo operation (n % 2), which effectively checks whether the least significant bit of n is set to 1.
n is then divided by 2, effectively shifting its bits to the right.
Approach-4 Summary :
In this, we are directly calculating Number of 1 Bits by using GCC inbuilt function (__builtin_popcount) in C++ and Integer.bitCount(n) in JAVA.
╔═╦╗╔╦╗╔═╦═╦╦╦╦╗╔═╗
║╚╣║║║╚╣╚╣╔╣╔╣║╚╣═╣
╠╗║╚╝║║╠╗║╚╣║║║║║═╣
╚═╩══╩═╩═╩═╩╝╚╩═╩═╝
✨ Timelines✨
00:00 - Introduction
#coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge#leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview#interviewtips #interviewpreparation #interview_ds_algo #hinglish #github #design #data #google #video #instagram #facebook #leetcode #computerscience #leetcodesolutions #leetcodequestionandanswers #code #learning #dsalgo #dsa