Lecture 7: LeetCode Problem Solving Session

[@himadridas8221]

When this man says, "samajh mein aa geya?, Chalo phirse samajhte hain", it gives a different kind of satisfaction ❤

[@shubh1373]

I have an doubt in first question that we had not defined any INT_MAX or INT_MIN so how can problem figure out meaning of these two some one please help

[@himadridas8221]

​@shubh1373 I am trying to understand your problem...INT_MAX or INT_MIN are not defined so how the code is being processed right ? These are called macros ...they initially hold the minimum or maximum values of an integer.. these are predefined.

[@shubh1373]

@@himadridas8221 thanks a lot bro cleared my doubt

[@BadalJena-oh6qu]

bro can you give a review of this course like there are other dsa resources like pep coding and pw. Is this course good enough.

[@himadridas8221]

@BadalJena-oh6qu look, I haven't completed this course, but I have watched some videos of this playlist, and came to know he is a fabulous teacher, who teaches you everything by holding your hand. So I would say if you want to learn DSA deeply you can follow this course, explanation will be top notch. Else if u want you can take his paid course of DSA which is yet to start

[@zaeemAtif]

Thank you bhaiya, don't know if you are even reading this comment or not, but just wanna say, the process you are teaching for solving quetions is waayyy more important and efficient than any course or youtube video I have watched. I definitely believe, this series is going to be a game changer.

[@tripuraridev5415]

yes you are right.....

[@nimrahasghar8955]

1000% right

[@hilariouschannel322]

Kitne vedios kr chuke ho xaiem bhai ?😅 Kaif

[@newhope5729]

bitwiseComplement(int n) { int nob=0,k=n; while(k!=0){ k=k>>1; nob++; } if(nob==0) return 0; int t=32-nob; k=INT_MAX^n; int l=INT_MAX>>t; n=l&&k; return n; Bhai isme kya prblm hai..... Question 2 ka code..... YE answer galat de rha par dry run karke dekho to sahi deta answer

[@newhope5729]

bool isPowerOfTwo(int n) { if(n==1) return true; int m=32; unsigned long k=pow(2,31); n=~n; while(m!=0) { n=n>>1; n=n|k; if(n==(INT_MAX-1)) return true; m--; } return false; } Bhaiya this is question 3.... What's the error?? It is giving false on problem 3.......

[@visheshkaran7083]

Bhaiya seriously you are doing a lot of work for the community. I'll also contribute a lot like you in future and thanks thanks a lot!!!

[@newhope5729]

bitwiseComplement(int n) { int nob=0,k=n; while(k!=0){ k=k>>1; nob++; } if(nob==0) return 0; int t=32-nob; k=INT_MAX^n; int l=INT_MAX>>t; n=l&&k; return n; Bhai isme kya prblm hai..... Question 2 ka code..... YE answer galat de rha par dry run karke dekho to sahi deta answer

[@newhope5729]

@rohitshaw-vd9lq Chhod diya bhai discontinue kar diya mai

[@newhope5729]

@rohitshaw-vd9lq Kuchh na bhai mai core electronics kar rha becoz iska abhi itna jaldi sab prepare ni ho payega iske liye 6 mahina bhar chahiye hota hai na mere ko time ni milta tha

[@ankitdaidanka9165]

I have done the power of 2 question using bit and right shift it was much easier instead. First question till now in this series which built my confidence as I was able to think out of the box. Thanks.

[@newhope5729]

bool isPowerOfTwo(int n) { if(n==1) return true; int m=32; unsigned long k=pow(2,31); n=~n; while(m!=0) { n=n>>1; n=n|k; if(n==(INT_MAX-1)) return true; m--; } return false; } Bhaiya this is question 3.... What's the error?? It is giving false on problem 3.......

[@aryyann05]

can u provide the solutiuon?

[@anuradhajadon4827]

Marked my attendance. loving the content.

[@apurvaparanjape8940]

Can u tell me how to mark the attendance

[@shubhambansal1495]

@@apurvaparanjape8940 video ko like karke

[@apurvaparanjape8940]

@@shubhambansal1495 ok thanks

[@rameshmalhotra9525]

kzbin.info/www/bejne/qnK0en6bZbp6fpY

[@IT-gw9rf]

Mujhse homework questions kyu nhi hote yrr😓

[@rahulkumarsingh7615]

In the last question of power of 2, U can also use bit property...........If(n & n-1 ==0) then it is a power of 2..

[@musicx5856]

Can you explain !!

[@shivu848]

Loved it bhaiya . We need more videos of leetcode questions discussion. I am a beginner in leetcode. By seeing your video i got the insight of how to approach a problem in leetcode. you are doing a great work..🙌

[@newhope5729]

bool isPowerOfTwo(int n) { if(n==1) return true; int m=32; unsigned long k=pow(2,31); n=~n; while(m!=0) { n=n>>1; n=n|k; if(n==(INT_MAX-1)) return true; m--; } return false; } Bhaiya this is question 3.... What's the error?? It is giving false on problem 3.......

[@RajatAswani]

ENJOYED!!! Being Honest, Quite fast, but the way he revises everything after submitting, it helps a lot and if you still don't get the concept, watch it repeatedly 2-3 times, you''ll get it. Completed on 10-11-2022 (21:22 pm).

[@shubhamuniyal9047]

MOST NEEDED VIDEO BROTHER❤️, WITHOUT DOING ACTUAL CODING, ONE CANNOT LEARN MUCH FROM IT :)

[@fatimasohail3027]

@shubhamuniyal9047 Why we write ans>int-Max/10?

[@PrakashKumar-xf7qf]

@@fatimasohail3027 because it goes beyond the last limit.

[@khushijain7689]

You explain that range part of problem in such easy words 🔥🔥

[@deepanshusaini353]

Maza aa rha hai lekin mujhse ho nhi rhe

[@Tej014]

Day : 7 attendance mark

[@sarcasmaursocial7331]

day 2

[@anishgujja3669]

keep going buddy

[@anishgujja3669]

Bro I have few doubts. Can we make a group or something like that to discuss our doubts. Share your insta I'd or something or just reply to this msg anyone who is reading

[@raunakbaliyan9918]

@@anishgujja3669 yes bro sure

[@GUNGUNLASTNAME]

​@@anishgujja3669 we will make telegram group for discussion

[@moutamarakshit]

bhaiya, i did the complement one with XOR :) mask = 0000....111 and then int ans = n ^ mask; kitna achchha laga jab sahi hua haha :)

[@AMIT-bf2id]

POWER OF TWO I have used binary digits to find solution , in binary number system 1 = 01 2 =10 4 = 100 8 = 1000 16 = 10000 My approach is I have converted n into binary string and checked if 1 appear excatly once public boolean isPowerOfTwo(int n) { int count = 0; if(n == 1) return true; if(n < 0) return false; String bin = Integer.toBinaryString(n); // int h = Integer.parseInt(bin); for(int i=0 ;i< bin.length();i++) { if(bin.charAt(i) == '1') count++; } if(count == 1) return true; else return false; } }``

[@anshbhatia9823]

Here instead of checking count for 1, you can also check that no bit until the first is 1 and return false immediately if so(by checking last bit and then performing right shift until n becomes just 1)

[@itzmranonymous]

@@anshbhatia9823 yup nice logic 🔥, but then too we will have to check whether solution is equal to 1 or not.

[@scientistakasha7293]

Ansh bhatia...you meant that start from the right most bit then decrement i?? If yes then yeah it will take lesser iterations and program will be faster

[@probably_yj]

My approach to the power of two question . I used the remainder of the division to check if the number is a power of two or not. Any Number which is power of two will always return n%2 = 0 till the end (except 0 and 1). So i ran a while loop till remainder is 0 . if we get remainder 0 till end then its power of 2 , else if we dont get 0 as remainder at any iteration then it will return false.. BTW , aapke videos bahut acche h Babbar bhaiya. Itne time se padhai me mazza nahi aaya tha mujhe :) class Solution { public: bool isPowerOfTwo(int n) { bool result ; while(n!=0){ int rem = n%2; if(rem==0){ n = n/2; } else if(rem!=0){ result = false; break; } result = true; } if(n==0){ result = false; } if(n == 1){ result = true; } return result; } };

[@debjitghorai9503]

easier way to do complement of base 10 integer:- (let n = 5); int num = 0; int i = 0; while (n!=0) { int bit,rev; bit = n & 1; // finding each bit of 5 i.e. 101 rev = bit ^ 1; // taking xor of each bit with one, which turns it into complementary (1xor1=0. & 0xor 1 = 1) n = n>>1; if (rev == 1) { num = pow(2,i)*rev + num; } i = i + 1; } return num; the xor operator turns 101(binary of 5) to 010(complement). and the num gives us output = 2;

[@piupiu634]

Nah bro .. even i thought abt this.. but this thing will count the 1s too..which aarent required... and then the result will be very big number

[@aashishrangdal9614]

Loving The Content🔥🔥 Babbar Bhaiya Is Getting Blessed Of Students Wishesh😍

[@vikashchoudhary7666]

Day 4 after starting first video, I'm fan of your teaching. thanks alot for this course. Right now I'm preparing for relevel DSA and I'm quite confident now at the end i can give my 100% learning in test

[@revantsinha1672]

this is my code for last question . bahut jayad simple simple socha😅😅 class Solution { public: bool isPowerOfTwo(int n) { bool istrue = true; if(n==1){ return true; } if(n

[@aryanisgood]

the last question can also be solved using this algorithm: all numbers that are powers of 2, have only one '1'in their binary form. eg 4 has 100, 32 has 10000, 8 has 1000. so we can run a loop and check the number of 1's in the number, if count ==1, then it is a power of 2 as '1' has appeared only once

[@sparshgamer1582]

Can you provide the code?

[@sameershaik5324]

We can also do power of 2 using log concept class Solution { public: bool isPowerOfTwo(int n) { if(n

[@youcube1]

😮😮

[@HarshKumar-mx9nj]

alternate approach for power of 2 question : #include using namespace std; int main (){ int n ; cout > n ; int rem = 0 , m = n ; for(;n>1;){ rem = rem + (n%2) ; n = n / 2 ; } if(rem == 0 || m == 1){ cout

[@yashsinghal3136]

3).Power of Two as said in the video there is another way we can solve this problem that i saw in discussion of leetcode that is to And(&) bit of the given number(n) and the bit of its previous number(n-1) and if they give 0 output then that number is in power of two. Example- Case1: n=2 2 in bit format=010 and(2-1=1) in bit format =001 when we And both the bits we get =000 Case2: n=8 8 in bit format=1000 (8-1=7) in bit format=0111 when we And them it will be=000 Similarly for all the cases CODE: if(n

[@AviSanX]

that was really beautiful and very smart, I also read the intial lines of the explanation of this code in discussion section but I was scared then and there. When I read yuor explanation, I understood it quite well and without any problem. Thanks.

[@divyampatni3571]

basically you are saying that only 1 bit out of 32 bits can be 1, to be a 2's power

[@saswatrath4646]

only if the left most bit could be 1 and rest all bit as 0 @@divyampatni3571

[@aleatoire6110]

well the complement can also be done in a simple way but the solution of bhaiya is also pretty good class Solution { public: int bitwiseComplement(int n) { int ans=0,i=0; if(n!=0) { while(n!=0) { int bit = n&1; if(bit==1) bit=0; else bit=1; ans = ans + bit*pow(2,i); n = n>>1; i++; } return ans; } else return 1; return 0; } };

[@RupeshKumar-ov5rt]

Thank you for all that you do. Your hard work doesn't go unnoticed. You're appreciated! thanks bhaiya 🥰

[@educulture1004]

36:23 qn can also be done using following logic: Since the given number would be a power of 2 so, its binary would be containing only a single bit as 1 and other as 0. Hence we can count the number of set bits in the number If count is 1 then true Otherwise false.

[@Scientest2004]

Bhut maja aaya bhaiya Content badiya he bhut Aap esse hi questions example or homework me dye na Or khi bhi koi bhi Kami nhi thi Full out full focus tha aapkye lecture me Or lykin 1gante ki bajaye muje 2h+30m lag gye Or Aaj phali bar leetcode per question fod kar kafi maja aaya Me ab se phale se or aagye future me abhi aapse yu hi judi rhungi bhaiya Thanks bhaiya etana badiya content hame free me sikhne kye liye . Fee kye rup me God ji se kafi sari badiya badiya wishes aapkye liye magu gi . Or muje bhi aapki hi trh ak badi company me interview fodna he . Jariya (way of path ) muje mil hi gya aapse padh kare . Bas ab entjar ak bar me hi select hokar dikhana he Microsoft kye interview me. Interview me select hone kye bade aati hu bhaiya aapse milne . Placement achi Kashi rhi thi aapkye liye apple 🍎 ka ipade bhi laugi .mari trf se gift samje kar rhkh lye ne aap vaise bhi aapko koi gift nhi dye ta (aapne ak video me bola tha esliye muje yade tha )esliye me dugi aapko sabse acha vala gift . Thank you bhaiya. Miltye he aapse bhut jald 😃😃😃👋

[@khushinegi3217]

Ques 3: bool isPoweroftwo(int n) { //It could also be like this if(n==0) { return false; } while( n != 1) { if(n % 2 != 0) { return false; } n = n/2; } return true; }

[@md.muhaiminulislam3611]

Bhaiya, I am from Bangladesh. I am in first year of my varsity and really I always enjoy your content. Your contents are so much helpful for me. I am following your guidelines and practising hard and soul to get internships as well as placements in future. Keep me in your prayers.

[@mlkgpta2869]

One of the best thing is that you know how students make mistakes. Before seeing your solution, all the tried methods by me was also shown by you, and it feels good when someone understand where students make mistakes. Great work, you not only teaches us what is the right path but also how to think, what is it the intuition. Aur aapke suggestions are too great, leetcode ka discussion is phodu.

[@sarthaktrivedi8180]

In that complement of an integer question, what I did was take the binary form of the actual number. Initialised a variable mask=1 and wrote the following code:- while((binum & mask)!= binum) { mask

[@ansarisufiyan4210]

Thank you so much bhaiya, at first when I started DSA, I thought that it is not for me, but now I am starting to have confidence in myself thanks to you.

[@adityamangla4794]

14:56, how come you used 'INT_MIN' and 'INT_MAX' without first defining those variables?

[@javacoder5351]

check this as well www.youtube.com/@maksrane100

[@rana.safikhan]

They are pre-defined

[@AdityaKumar-xk8fk]

because it is predefined integer class

[@RohitKumar-oo8lo]

BOHOT zyada mazzaaa aa rha hai bhaiya, sab samajh aa rha hai...Aap plz time pe khatam kar dena iss course ko 🙏 Amazon crack karna hai ab to kisi bhi haalat me!!

[@vedantshivarkar6057]

Didi you complete the course??

[@fgh427]

Greatest teacher of all time... No one... literally no one is even 1% close to you..

[@aasthatripathi6952]

Lecture 7 ✅ Maza aya 👍

[@Rohitkumar-kg1bh]

class Solution { public: bool isPowerOfTwo(int n) { int count = 0; while (n != 0) { if ((n & 1)==1) { count++; if (count > 1) break; } n = n >> 1; } if (count == 1) return true; return false; } SIR aapne bits padhaya to usi direction me soch kar ye code lagaya 15 mins se v kam lage sochne me , or jab code accept hua to bahut achha feel hua ThankYou❤

[@vinitpatil4137]

Bina paise ka itna badiya course, u are a legend. Udemy vale pe tik hi nhi pata hu aur expensive course afford nhi kar pata, itna acha koi bachpan me sikhata to maza aa jata. All the best, keep up the good work. Way of teaching is too good no doubt.

[@manukhurana97]

Simple 3 line java Sol: 17:40 if(n==0) return 1; int i = 0, m = 0; while(n != 0){ m += Math.pow(2, i++) * ((n & 1) == 0 ? 1 : 0); n >>= 1; } return m;

[@prajwalurkude007]

I able to solve 2 out of 3 problems with different approaches simultaneously with sir in the video!

[@pritishanand]

*2 out of 3

[@prajwalurkude007]

@@pritishanand yea corrected

[@fukedup1028]

great

[@aavyasingh6489]

I study from so many channel ,but no one can solve a question on this level . When solving these minded question ,it exercise our brain and buildup our concept also . THANKS A LOT BHAIYA

[@AmitSharma-nv2oj]

Can you tell how he can use int max and int main without declaring them

[@deadlyfury6317]

@@AmitSharma-nv2oj maximum range if already defined for integers... so we can use directly

[@crowdmedia7815]

bhaiya you r such an incredible teacher who selflessly trying this hard !

[@CodeCommitC]

Another Alternative Of Power Of Two solution in Javascript: Explanation: if we see any number which is a power of two in binary format there would be only single binary bit which is 1 other must be zero. so we can simply check the count of binary 1's in the number, if it is 1 in total it means the number is power of two otherwise not , and we have to check for the edge case as well which is when the number is 0 var isPowerOfTwo = function(n) { let count=0; if(n==0) return false; while(n!==0) { let digit = n&1; if(digit) count++; if(count>1) return false; n >>=1; } return true; }; Thank you so much bhiya for this amazing series, lot of respect

[@aritralahiri8321]

Very much premium content . Thanks brother really appreciate the hard work .

[@SumitSingh-wz3wt]

This also worked int main(n){ int x=0, temp=n; while (temp>1){ temp /= 2; x++; } return pow(2,x) == n; }

[@gaurijadhav9161]

for the first time i solved the 2nd problem by myself...literally soo Happy..Amazing Content!!

[@user-qr1xq6cq5u]

Bhai party chahiye

[@pragyanshandilya4015]

Hello Babbar bhaiya recently saw your stl video, so tried that complement question with the help of vectors class Solution { public: int bitwiseComplement(int n) { int bit,binary=0,i=0; vectorv; if(n==0) return 1; else { while(n!=0) { bit=n&1; binary=bit*pow(10,i)+binary; n=n>>1; i++; } while(binary!=0) { bit=binary%10; v.push_back(bit); binary=binary/10; } reverse(v.begin(), v.end()); for(auto&it:v) { if(it==1) it=0; else if(it==0) it=1; } reverse(v.begin(), v.end()); int sum=0,s,k=0; for(auto&it:v) { s=it%10; sum=sum+s*pow(2,k); k++; } return sum; }} };

[@only_for_fun1234r]

Bhai tum god ho ky??🥺🥲

[@pragyanshandilya4015]

@@only_for_fun1234r kya hua bro galat hai kya ;(

[@h_a_r_s_h01]

Bhaiya if possible can we get solution for the homework of the previous lecture + Motivate us(can be of 15-20 seconds) at the end of every lecture if possible ? " YOU'RE BEST FROM THE REST ❤️ "

[@Incurabl]

Try more n more to do by yourself or in last Googling will help for sure.u can learn more this way

[@Captain_Pikachu]

41:45 is question ko hm aese bhi kr skte h , agr koi number 2 ki power m express hoga to uske binary m exactly ek bar 1 aayega to hm count variable initialise kr skte h aur number ko right shift krke check kr skte h kitni bar 1 h , code: class Solution { public: bool isPowerOfTwo(int n) { if (n>1; } if (count==1){ return true; } else{ return false;} } }; // n0 (khud kiya h aur maja aaya 😎😎) bhardyash

[@priyanshpatel2078]

Happy to comment it i know i am late but i started this playlist before 2 to 3 days i just give my answer of 2nd Question - 1's Complement of number With My Explanation //Complement Of Base 10 Integer /* The complement of an integer is the integer you get when you flip all the 0's to 1 and all the 1's to 0 in the binary representation for example - The integer 5 is "101" in binary representation and its complement is "010" which is the integer 2 in Base 10 Example - Input n = 5 Output - 2 */ #include #include using namespace std; int main() { int number; cout

[@architbatham4261]

Bhaiya aap bohot asan bohot asan bol k question ko sahi me asan kr dete ho

[@strivermonk9424]

Alternate solution of complement of Number using XOR Operation : class Solution { public: int findComplement(int num) { int res = num; int mask=0; while(num!=0) { num=num>>1; mask=(mask

[@ghanashyam7701]

Bro but the return value res^mask is in binary form right !?

[@vijaymathe8346]

41:37 in this question we can also play with the binary representation of the number here is the two liner code... if(n

[@rahul.murari]

I have 9 year experience in IT and following you from last couple of months, your content is awesome, I believe everyone surely will achieve something from here and me also, Thanks, Kudos. Love ❤️

[@AmitSharma-nv2oj]

Can you tell how he can use int max and int main without declaring them

[@random-0]

@@AmitSharma-nv2oj in c++ these are inbuilt variables

[@miaalora5707]

Bht mza aa rha h pdne ap bhaiya bhtt acha pda rhe h

[@neerajslab]

Bhaiya in first question why r u dividing by 10 in INT_MAX and INT_MIN? can we directly write the condition like this :- if(ans > INT_MAX || ans < INT_MIN) ?

[@samialvi4226]

Mera b yhi question h apko ager mila ho iska solution to please let me know

[@onkarswami]

To avoid termination of program due to overflow.

[@pragatijain8261]

Reach++++ Best ever DSA series...Thanks bhaiya

[@adityamangla4794]

At 15:00, shouldn't we use 'break' so that 'while' loop can be exited if we exceed the range?

[@khatariinsaan5284]

no need to break it will just return the answer

[@rajneeshkhare6106]

Power of 2: Another approach - since all numbers in powers of two have exactly one number of one's in their binary representation, we can solve it by performing and bitwise operation and counting the number of one's. class Solution { public: bool isPowerOfTwo(int n) { int count=0; if(n>1; } if(count==1) return true; return false; } };

[@h_a_r_s_h01]

Bhaiya if possible can you repeat the concepts of 7th lecture in future, actually today's lecture was quite barely understandable (especially INT_MAX/10 and INT_MIN/10 + Last Question i.e. Power of 2). Kindly take it as feedback. If possible please provide couple of questions on these concepts.

[@TOMGAMING-hy9hi]

yes it was very difficult logic......

[@AjaySingh-ty2wm]

bro watch it again , i also didnt got in first attempt . now its clear

[@arungound2011]

@@AjaySingh-ty2wm Can u solve my doubt at 15:02 Instead of ll why we didn't use && ...both condition are true right??

[@word___addict8768]

@@arungound2011 bro both conditions to kabhi true hogi hi nhi na to if ka fayda kya hua

[@retro9511]

@@word___addict8768 bro can you tell me what is INT_MAX ?

[@abhijeetsharma7858]

Bhaiya this is also an approach for powerOfTwo problem bool powerOfTwo(int n){ if(n==0){ return false; } while(n%2==0){ n/=2; } if(n==1){ return true; } return false; }

[@abhishekagarwal4671]

I have a doubt on Integer Reverse Question In the last iteration if ans is getting greater than INT_MAX after adding digit and in last iteration x will be 0 and it does not get into loop to check the condition...will the code work fine please explain if somebody can get my point

[@masterquiz2307]

17:35 My approach ...SHOWING 100% FASTER class Solution { public: int bitwiseComplement(int n) { int p=2; while(p

[@rohan3944]

Day-1 /. 3 dec

[@sakshamgupta7503]

even i started

[@rishurk9955]

2nd question different approach- public int bitwiseComplement(int m) { int mask = 0; int n = m; if(n==0){ return 1; } int i = 0; int ans = 0; while (n!=0){ int bit = (n&1)^1; ans = (int) (Math.pow(2,i++)*bit)+ans; n=n>>1; } return ans; }

[@AbdullaVlogs]

Thanks, Bubbor Bhyya Getting too much involved while solving questions. I feel like my mind is now open too much to accept any test cases in 0ms hahaha😀

[@sonumondal7798]

I have explored so many DSA courses on youtube but this one is next level ........No one can come close to this , I am loving it

[@user-mz9yl1vn8u]

going to college and watched your videos about dsa college is not opened yet but i started my preparation by watching your videos and are very helpfull sir

[@nausheenkhan5896]

Thanks Bhaiyya..... feels so good that after watching your series, I am hopeful that I will be able to get a good grasp on data structure which i have been trying for years...Thanks a lot for giing this to community

[@ahmadnadeemjb]

Best Video On this topic on KZbin

[@aniketnavele1131]

bro kuch nhi samaja

[@Saalim_Malik]

it can also be done with the help of bool for answering 'yes' or 'no' which I found easier... ////check if the no. can be expressed in the power of 2 int n; coutn; bool b = 1; for (int i = 0; i < 31; i++) { int ans = pow(2,i); if (ans == n) { b = 0; break; } } if (b == 1) { cout

[@yashaditya6179]

instead of doing (~n) & mask we can also do n ^ mask . ' Also thanks a lot bhaiya , this series is immensely helpful !

[@pawanbhatt314]

thankyou for replying, i was hella confused coz I did (n ^ mask)

[@VishalKumar-bj4nl]

I know it's a bit long but it is understandable as he taught this one in his next lectures. // Starts from here. int count = 0; while(n != 0){ int bits = n & 1; if(n < 1) return false; else if(bits == 1) count++; n = n >> 1; } if(count == 1) return true; else return false; It simply states that if there only one '1' in the binary form of any input except negative no's then it will increase the count variable for one single time only otherwise returns false. Thank you.

[@mohammedjunaidahmed9821]

I don't know C++. I learned dart and was planning to go for DSA using Java. Could you make this course inclusive for Java programmers as well? Otherwise we have to learn C++ also

[@Khan67387]

you can also do the 3rd question by using while loop class Solution { public: bool isPowerOfTwo(int n) { int i=0; while (i

[@gourabdas8430]

Sir bohot maza ah raha h…bohot jada..sir m Mechanical background se hu….bt..aapka class kar k….m bohot sara chiz sikh giya hu.…khud se problem solved kar pa rah hu Thank u so much sir…🙏🙏🙏🙏❤️

[@mrittikadey3734]

Vaiya awesome ..from Bangladesh 🇧🇩..I saw a lot of videos before but most of them trying to tech us how to code but you teach us how to think ,how to understand.. Respect bro..

[@prithvirajpatil6802]

void reverse(int n) { while (n != 0) { int rev = n % 10; n = n / 10; cout

[@priyapandey8951]

@Babbar Bhaiya, Power of two hum ase bhi karskte:- jitne bhi numbers 2 ke power me hai unka binary number asa hai ki usme sare bits 0 hai except ek bit joki 1 hai eg. 4=100, 8=1000,16=10000 and so on. To hum srf wo ek bit ke liye check karenge aur count variable lga kar karliya. class Solution { public: bool isPowerOfTwo(int n) { if(n>1; } if(count==1) return true; return false; } };

[@_Sourav-us4ph]

//Complement class Solution { public: int bitwiseComplement(int n) { if(n==0){ return 1; } int c=(int)(log(n)/log(2)); return (~n)&(int)(pow(2,c)-1); } }; //Power of 2 class Solution { public: bool isPowerOfTwo(int n) { if(n

[@shubhamKumar-jg8ku]

Thank you bhaiya ,kbhi socha nhi tha ki coding itni achi lgegi ,ye sirf aapke pdhane ke trike se hi ho paya h,kya smjhate ho awesome.thank you so much for this series.

[@_SouravRana]

#include using namespace std; int main() { int n=64; int ans=1; while(ans

[@yashrajgharat10]

Easy way to find a compliment of a number by using the concept of binary to decimal & decimal to binary conversion int main() { int n,ans=0; coutn; int i=0; if(n==0) ans=1;//return ans =1; while(n!=0) { int bit=n&1; if(bit==1) bit=0; else bit=1; if(bit==1) ans=ans+pow(2,i); n=n>>1; i++; } cout

[@aryansharma144]

Best DSA content in KZbin ❤️🥇🔥

[@khushiupadhyay5193]

sir i have solved the power of 2 problem like this: bool isPowerOfTwo(int n) { int i=0; while(pow(2,i)

[@saqihere]

Amazing Session:)

[@shreyansh_yadav]

Bhaiya last question mei ek or approach kr skte hain ki log2 le lein dono side.. n == 2^x log2(n) == x Yahan pr check krlein ki log2(n) agr sahi mei perfect integer hai to , sahi mei power of 2 hai agr log2(n) iski value float ya double mei hai to nahi hai... code : bool isPowerOfTwo(int n) { double m = log2(n); if(n == 0){ return false; } else if(floor(m) == m){ return true; } return false; }

[@divyadeep7679]

finally....m no more worried about DSA...just have to continue watching this

[@ayushkumarjha6870]

power of two bool isPowerOfTwo(int n) { if(n1){ if(n%2==1){ return false; } n/=2; } return true; }

[@tanishsolanki9277]

class Solution { public: bool isPowerOfTwo(int n) { for(int i = 0;i

[@chandan.cg799]

int bitwiseComplement(int n) { if(n==0){ return 1; } int temp = n; int bits = 0; while(n){ n=n/2; bits++; } int number = 1

[@ashishmishra155]

while(n > 1){ n = n / 2; } return n == 1; sirf itna sa code the last problem ka easy bhi hai bht aur efficient bhi

[@GamingandTechChannel]

Problem Power of 2 in js let n=4; while(n>1){ n=n/2; } return n===1;

[@priyasharma8315]

Javascript Solution : 1.Reverse Integer var reverse = function(x) { let ans = 0; while(x !== 0){ let digit = parseInt(x % 10); ans = (ans * 10) + digit; if (ans < Math.pow(-2, 31) || ans > Math.pow(2, 31) - 1) return 0 x = parseInt(x/10); } return ans; };

[@ritishasingh5098]

Another approach to Base 10 Complement class Solution { public: int bitwiseComplement(int n) { if(n==0){ return 1; } int ans=0,i=0; while(n!=0){ int bit=n&1; if(!bit){ ans=pow(2,i)+ans; } ++i; n=n>>1; } return ans; } };

[@imgroot9102]

hello bhaiya maine complement wala question usse bhi simple solution kiya wo bhi khud se kafi proud feel ho rha h ki going in the right direction.. class Solution { public: int bitwiseComplement(int n) { int i=0,ans=0,val=0; if(n==0){ val=1; } for(;n!=0;n=n/2){ int dig=n%2; if(dig==1){ dig==0; i++; } else{ dig=1; val+=pow(2,i); i++; } } return val; } };

[@rpgaming757]

Thanku you bhaiya ,Bhaiya seriously you are doing a lot of work for the community. I'll also contribute a lot like you in future and thanks thanks a lot!!!

[@kashishaggarwal4756]

You explain each and every thing in a very nice manner ..... Thank you so much for your efforts brother

[@pulkitgarg7021]

int bitwiseComplement(int n) { // edge case if (n == 0) return 1; int i = 0; int num = 0; while(n!=0) { int bit = !(n&1); num = (pow(2, i) * bit) + num; n = n >> 1; i++; } return num; } This is also a possible solution.