201. Bitwise AND of Numbers Range | Remove Last Set Bit | Bit Manipulation

[@rajrajesh1669]

1:20 😂 bro that cracked me up...

[@Sakshi-yq4jm]

this was literally the best explanation, i have been trying to understand the reason behind this approach for quite some time now, been surfing youtube all day today. and finally. good work. thank you!

[@ARYANMITTAL]

Glad it was helpful Sakshi 🙇🫡🫂

[@aayushyavajpayee4989]

from 8:16 to 8:48, you can avoid that and video will become shorter and crisp. You are doing great

[@tusharkumarraj6066]

loved the explanation. Watched explanation video from neetcode and didnt understand much , this one was so simple and intiutive

[@sameershah1492]

Amazing approach ❤

[@ThisIsNotMyYoutubeAccount]

Can you make a video of your day time table ?

[@kushal8261]

Yes please!

[@Sahilsharma-sk5vr]

yes please

[@ganeshsharma5810]

amazing approach bhaiya

[@abhinavdubey_7]

Thankyou my brother amazing.

[@shalinijha7359]

just thankyou for existinggg

[@learningmaths786]

Thanks Aryan ❤

[@HarshSharma-hi9vc]

nice explanation

[@codedecode8686]

nice approach

[@CodeMode9313]

badia

[@alifrahman7099]

thanks

[@ritishrai581]

Oh f . I did this using the BIT MANIP trick you taught in one of your previous videos . I guess the tc would still be equal to the no of bits right? #define ll long long class Solution { public: void calcBitRange(vector&bits, int num) { if (num==0) { return ; } if (num==1) { bits[0]++ ; return; } else if (num==2) { bits[0]++; bits[1]++ ; return ; } ll bitLen = log2(num); ll nearPowerOf2 = 1ll =0 ;i --) { bits[i]+= nearPowerOf2 >> 1ll ; } calcBitRange(bits,num-nearPowerOf2); return ; } int rangeBitwiseAnd(int left, int right) { vectorrange1(65,0) ; vectorrange2(65,0) ; if (left==0) return 0 ; if (left>1) calcBitRange(range1,left-1); calcBitRange(range2,right); ll ans = 0 ; for (int i =0 ; i

[@whoshyam]

please provide your onenote page link

[@parthbhatti4151]

why chatgpt says O(log(right)) time complextiy ?

[@sahilkarwani9629]

if u take the maximum value of right according to question then it will be 2^32 then O(log(2^32)) gives you O(32) which is O(1).

[@RachitKumar-yp2ub]

worst case time complexity is O(32). cause at max we might end up taking 32 traversals.