Power of 2 | Leetcode

[@yjj410]

I did it by droping the least significant bit and checking if the number becomes zero. To drop least significant bit i used (x & (x -1)) expression.

[@techdose4u]

Good :)

[@siddharthsingh9409]

I also did the same!

[@agileprogramming7463]

Crystal clear explanation as always sir! The efforts you are taking is remarkable

[@techdose4u]

Thanks agile :)

[@harinaaraayans3453]

we can just check if the number is a divisor of the max value of that programming language.. example... in C++ if a number divides 32768, it i s a power of 2. this is much faster than taking log

[@techdose4u]

Use built-in function instead.

[@CSKAASIPRASANTHA]

Excellent explanation with efficient solution.

[@acxd]

another approach (using in-built function) if(__builtin_popcount(n) ==1 and n>0 ) return true; else return false; the above is also an O(1) approach

[@techdose4u]

Yes....built-in is O(1).

[@abhishekdas8313]

Language?

[@paranjoypaul3329]

One more method... if(n

[@techdose4u]

Everyone has used the same approach 😅

[@jayajmire6779]

What is its run time

[@rohanprak]

@@jayajmire6779 O(1)

[@rohanprak]

@@techdose4u indeed , bro , everyone who knew fenwick tree, may have used it probably, as Fenwick is the game of last set bit. XD

[@chandanchandak5194]

(n &(n-1))==0 is enough to determine if its power of two or not

[@mujahidpasha4440]

Really appreciate your efforts and time you spend on teaching... Thank you very much... Saw a few videos of you and right away subscribed to your channel... It's worth subscribing :)

[@techdose4u]

Thanks :)

[@AbhinavKumar-dr4ef]

Krishna bless you. Nice explanation.

[@sudhanshukumar1558]

In python I did Exponent = int(math.log(n,2)) return True if 2**Exponent == n else False Thanks for sharing your solutions, the 2 and 3 solutions were great 😃

[@techdose4u]

👍

[@ayonsinha2075]

time compjlexity will be log n for this case ...isn't it??

[@shivanshutiwari5646]

I loved 4 rh method. Its great

[@techdose4u]

👍

[@bhanuprasad6793]

nice explation worth of watching

[@karansagar7870]

Nice approach

[@techdose4u]

👍

[@lapujain]

Amazing teaching skills, you made me learn a new way to do problems in a very simple way. My solution to this problem was using bitset class P.S : Without sizeof(int) it will pass 1107 test cases. class Solution { public: bool isPowerOfTwo(int n) { bitset i(n); if (i.count() == 1) return true; return false; } };

[@techdose4u]

Nice :)

[@saisrisai9649]

Awesome 👏🏻 👏🏻👏🏻

[@techdose4u]

Thanks :)

[@nightfury1250]

Here the code of all four methods in javascript. 1) By Recursion const isPower2 = (n) => { if (n === 1) return true; if (n % 2 !== 0) return false; return isPower2(n / 2); }; 2) By Counting Set Bits const isPower2ByCountingSetBits = (n) => { let count = 0; let add = 1; while (add

[@ShubhamMahawar_Dancer_Actor]

In some post i also saw solution if (n&(n-1))==0 return true else false but i think that will will fail for number out of range for int.??

[@techdose4u]

Number will be in int only. So no problem.

[@ShubhamMahawar_Dancer_Actor]

@@techdose4u but its giving runtime error while running this solution You can check once please?

[@rohanprak]

this will work even if no is out of range of Long ! , error kaha se aa gaya bro ? comment your code.

[@hakoHiyo271]

Very nice video of explanation! Thank you!

[@techdose4u]

Welcome :)

[@andriidanylov9453]

Brilliant. Appreciate.

[@fatimajaffal9843]

class Solution { public: bool isPowerOfTwo(int n) { if(n

[@AnandKumar-uy7nn]

Thanks for giving such a great content. 💗💗💗

[@uditagrawal6603]

n&n-1 = 0 for power of 2 with n > 1

[@techdose4u]

👍🏼

[@ManojKrVerma-vw4dx]

You have told that tc for log method is O(1) bt How? Finding log itself is log N. Correct me if I am wrong ??

[@techdose4u]

There are many ways to find log. You don't have to specifically use log2. But as far as I know, standard library log2() is O(1). You can use __builtin_clzl(N)-1 to find log in O(1). This is guaranteed. Follow stackoverflow for this.

[@karansagar7870]

@@techdose4u can u explain this in coming video just brief

[@blahita]

Thanks for the great explanation! What is the software that you are using for explaining?

[@Ayush-bj8gk]

sir in first method 3/2 will give 1 where as 2/2 will also give 1 How to handle this?

[@atulpunde4786]

Awesome..❤️

[@techdose4u]

Thanks :)

[@satyamdubey4110]

👏👏❤

[@techdose4u]

❤️

[@PARAxITACHI]

how 2nd method hv complexity og log2(n).

[@KD-xi9wu]

What if the number is too large and given in the form of String ?????????????????????????

[@techdose4u]

Just check that 1st char of strong should be 1 and rest all 0s. That's it.

[@Yash-uk8ib]

sir plzz make a video on time complexity analysis plzz..

[@techdose4u]

That will be possible only when the leetcode challenges ends.

[@HimanshuSingh-ob9qo]

👍

[@techdose4u]

👍

[@fb_a]

Isn't the internal implementation of log() function time complexity is O(log(N) )?

[@ayonsinha2075]

exactlly that is my confusion

[@Pratik_yadaw]

thanks you so much sir!

[@sahithikomirishetti3380]

Love in with your explanation including the complexities

[@techdose4u]

Thanks :)

[@JAG112AO]

what does it power of two mean?

[@ankitalone8906]

Thank you so much ❤️

[@01utpalraj95]

There is one more method that if n&(n-1) is zero then n is 2 ki power sm.

[@prabhatchanchal]

I used N>0 && (N&(N-1)==0)

[@techdose4u]

Nice

[@HimanshuKumar-bf3pq]

last method will fail for 5 test cases from almost 21000 test cases, one failed test case is 243 as log(243)/log(3) = 5.000000000000001 which fails ceil and floor

[@mayankprakash9651]

OMG Aaj hm bhul gaye the aur time ho gaya tha 12:27 am hmko laga gaye phir jakr dekhe tab 3 min bacha hua tha due to PCD time , uske baad hm yaha wala GitHub code copy kar diye jaldi se.😑

[@techdose4u]

Time ends at 12:30 AM noon in INDIA.

[@mayankprakash9651]

@@techdose4u Oo mistake am and pm.

[@techdose4u]

Try to solve on time bro. You have 24 hrs.

[@j4hnvi]

First Method is incorrect. If you take 14 as example and divide it by 2 we get 1 in the end but it isn't power of two.

[@DHEERENDRAKUMARBEE]

bhai 1 test case paas nahi ho raha

[@javi2082]

Check if x & (x-1) == 0

[@mayankmaurya8631]

(n>0 && ((n & (n-1)) == 0))

[@AbhishekKumar-sf6no]

n & n-1

[@siddharthasharma9316]

N &-(N)==N return true else return false

[@siddharthasharma9316]

Sorry I type wrong.....N &(-N)==N

[@siddharthasharma9316]

Where -N is 2's complement