Find rightmost different bit between 2 numbers

[@uditagrawal6603]

Instead of checking for equality of mask, we also do: 1. XOR of both the numbers. 2.Do mask with XOR value. 3.Search for the first occurence where XOR&mask == 1. WDYT?

[@shikhar2k01]

LGTM

[@patilsoham]

greatest explanation. everything looks simple when you explain it.

[@techdose4u]

Thanks :)

[@ManishKumar-ys5oj]

simple aproach: taking xor of both no...and find index of right most setbit of xor.

[@techdose4u]

Correct.

[@aryanjindal127]

sir thanks for your simple and different solutions bit masking is really really very helpful

[@techdose4u]

Yep 😊

[@motasim3456]

We can also do something like, right shifting both the numbers until one of them become even and another one becomes odd.

[@saichennu5288]

Great explanation sir..

[@techdose4u]

Thanks :)

[@AtulDislay]

Great explanation as usual 🙌

[@techdose4u]

Thanks :)

[@saichennu5288]

Sir, upload some dsa concepts like greedy, dynamic programming,brute force etc.

[@techdose4u]

Bro currently I am doing some videos on graph because many of you had requested it. I will be going home for Christmas and New year. So I will make some videos on hash and heap. After I come back I will surely continue DP and start with greedy. This limitation is due to time. I hope you understand :)

[@abhaytiwari6411]

bro I only understand bruteforce method thanks for explaning in easy way

[@techdose4u]

You dint understand the bitmasking method?

[@abhaytiwari6411]

Ya because it is complicated

[@HomemadeTechARC08]

Sir, the efficient approch is taking the same no of operations as bruteforce.

[@HomemadeTechARC08]

Still we are going to compare the N bits in the binary representation

[@techdose4u]

@@HomemadeTechARC08 Actually in the bruteforce, you need to find the binary representation but in efficient approach, we are solving using decimal numbers without converting it to binary :) So, time is very less as compared to bruteforce.

[@shubhiagarwal4047]

sir isme koi bhi glti nhi h but output is not coming by ur approach #include using namespace std; int main() { int t;cin>>t; while(t--) { int m,n;cin>>m>>n; int count,mask=1; while((mask&m)==(mask&n)) { mask

[@techdose4u]

For which test case is the answer coming wrong?

[@shubhiagarwal4047]

@@techdose4u 387 885 iske liye...

[@shubhiagarwal4047]

@@techdose4u sir really u replied. I m glad and relieved

[@ayushgarg5929]

the better solution will be firstly take xor of both the numbers then find out the position of first set bit in the resultant ...that will be our answer

[@priyankapardesiramachander871]

sir, can't we just XOR a and b.. then perform 'Find rightmost set bit' method?

[@techdose4u]

Yes

[@AshutoshJha-k1l]

Instead of this approach we can use the xor approach, by applying the concept that (0^1) or (1^0)=1

[@rjagan6951]

Corner case , what if both values are same??

[@susovanghosh6050]

I think it can even be done like this as well : answer=a xor b and then check the rightmost set bit of answer.

[@muktiranjanmohapatra9785]

🙏 Teacher

[@techdose4u]

:D

[@bhupeshpattanaik7150]

one more way , easier int firstSetBit(int n) // gives first position where 1 is found { int i=0; while(n) { i++; if(n&1) return i; n = n>>1; } } int posOfRightMostDiffBit(int m, int n) { if(m==n) return -1; int t = m^n; // finding XOR of both number return firstSetBit(t); }

[@jaydeepmahajan6598]

You can also find this taking square root of x & (-x) +1

[@b.sainathreddy4567]

#include using namespace std; //method 1 int main() { // taking the xor of two numbers and finding the right most significant bit // will give the right most different bit int a = 10, b = 14; int c = a ^ b; int rsbm = c & -c; int counter = 1; cout

[@abhilashpatel3036]

Didn't watch till end. But the efficient approach will be using XOR. His approach is too slow.

[@rajpalsinghgamer95]

No no😂🤣🤬👿🤪you are a kid🤘

[@jpakash1999]

sir please solve the question i sent you through facebook

[@techdose4u]

I think you sent codechef course. If I get time I will try to explain. I am busy this month.

[@jpakash1999]

Sure thank you anyways

[@jpakash1999]

That was a question asked in zoho interview sir ,don't know it's actual source