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 🙇🫡🫂

[@tusharkumarraj6066]

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

[@aayushyavajpayee4989]

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

[@akhilkarumanchi458]

Can you make a video of your day time table ?

[@kushal8261]

Yes please!

[@Sahilsharma-sk5vr]

yes please

[@sameershah1492]

Amazing approach ❤

[@ganeshsharma5810]

amazing approach bhaiya

[@abhinavdubey_7]

Thankyou my brother amazing.

[@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

[@learningmaths786]

Thanks Aryan ❤

[@shalinijha7359]

just thankyou for existinggg

[@codedecode8686]

nice approach

[@HarshSharma-hi9vc]

nice explanation

[@CodeMode9313]

badia

[@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.