3133. Minimum Array End | Bit Interweaving | Bit Interleaving | Bit Manipulation | 2 Pointers

In this video, I'll talk about how to solve Leetcode 3133. Minimum Array End | Bit Interweaving | Bit Interleaving | Bit Manipulation | 2 Pointers
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I am Aryan Mittal - A Software Engineer in Goldman Sachs, Speaker, Creator & Educator. During my free time, I create programming education content on this channel & also how to use that to grow :)
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[@user-hv7yx3ps5z]

The way you explain ideas is like watching a Bollywood movie; it's very engaging.

[@Noob_Coder1234]

best youtuber for leetcode , clean explainations with most optimzied codes !! god bless u bro!!

[@shivamgurjar8979]

so well explained

[@rajan-u6b]

love u sir, nice explanation.

[@InWonderland-z2l]

now i've watched your videos to the point that i hit like as soon as the video starts and then start watching it😂same as what i do with striver because i KNOW it's going to be epic🌟

[@mukulkhanna5071]

best video on entire earth

[@slayer8792]

This is a great solution , very nicely explained

[@rishavsaha5254]

Thank you so much for the video. There can be a easy solution for this as well: Increasing the unset bits: long long minEnd(int n, int x) { long long a = x; n--; while(n--) { a=(a+1)|x; } return a; }

[@akshatsingh6036]

Thanks for getting me a TLE ! Not optimal though

[@kashishchawla2754]

@@akshatsingh6036 it wont bro!

[@divyanshagrawal6267]

Cool Solution Dude :)

[@aafighters4535]

Keep going bro.

[@psk9736]

When setting xbit[i] and nbit[i] I used xbit[i] = x & (1 i) & 1 (n >> i) & 1

[@ARYANMITTAL]

Let us say you want to know existence of 2nd bit i.e for a number 4, which has binary representation as 100, then your code will say xbit[2] = 4 while, you wanted to know just that if that 2nd bit was set or not, which means you wanted xbit[2] = 1, xbit will have only 2 values 0 or 1, saying if some bit was set or not

[@psk9736]

​@@ARYANMITTALoh accha 😅😅 understood thankyou ❤❤

[@Danish-saifi1]

@@ARYANMITTAL int se & krne pr int me convert ho jata hai no? or 1 se krne pr binary me rehta hai?

[@ayushtandon1719]

You can also do this xbit[i]=x&1; x>>=1; nbit[i]=n&1; n>>=1;

[@lavanya_m01]

Awesome explanation bro!! Can you pls solve 3134 also?

[@prathamsharma4416]

nice 1

[@AdityaYadav-qf9qc]

@aryanmittal can we apply binary search on answer on this question??

[@hrishavraj7706]

👌

[@ias5462]

I understood the logic..but i didn't get the intuition why n-1 number only we have to do the 2 pointer thing..like is there any concept behind

[@bhushanasutkar6135]

Thanks

[@mrityunjoybarman9098]

Why my code is giving me wrong ans for 56 testcases ? public long minEnd(int n, int x) { int j = 0; long ans = (long)x; for(int i=0; i> i) & 1); if(bit == 1) continue; if((1 n-1) break; int newBit = (((n-1) >> j) & 1); j++; if(newBit == 0) continue; ans |= (newBit

[@RitikRanjan-fn9me]

Intuition ban gya tha , but implement ni hua :/

[@dhruvrawatt9]

I give up on coding

[@ayushmishra1207]

nBit[i] should be equal to ((n-1)>>1)&1 na instead of (n>>1)&1 as You have taught the same thing at 12:39

[@242deepak]

Easiest code: class Solution { public: long long minEnd(int n, int x) { long long ans=x; long long k=n-1; for(int i=0;i

[@pranjalverma8239]

bhai kuch samaj nhi aaya 😢

[@jimit2795]

Sahi me. Sab upar se gya.

[@panapatre9855]

while(until don't get it) { repeat_the_video() }

[@Anonymous____________A721]

Uff​@@panapatre9855