Google Interview Question | Bit Manipulation | Solution With Code | Interview Bit Problem

Hi,

This is the eighth video of our playlist named "InterviewBit Problems and Solutions" named as "Single Number II".

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[idk_]

I have searched a lot for good explanation...finally i got it...
Thanks bro.

[harinijeyaraman]

Good explanation ! Found this video really helpful !

[Mauryarohit]

Very nice and straightforward explanation. Thanks!

[AvinashSingh-kloz]

Hey man
Help me out
I tried your code
But if I change the 12th line of your code from '!=0' to '==1', the code stops working
If it's binary, it should be either 0 or 1 right?
So I should be able to do A[j] & (1<<i) == 1 as well
But it doesn't work for some reason and I am not able to figure out why it doesn't work

[subzeroOrkut]

Will this also work with negative numbers, e.g. what if the input is
[-2, -2, 1, 1, -3, 1, -3, -3, -4, -2]?

[mukulbakshi]

Please correct me if I'm wrong, like if the number that appears once and the number which appears multiple times has bit set to 1 then it will produce two output which is not divisible by 3 example
A = [5, 3, 3, 3]
binaryA = [0101, 0011, 0011, 0011]


Sum of First bits%3 = (1+1+1+1)%3 = 1
Sum of Second bits%3 = (1+1+1+0)%3 = 0
Sum of Third bits%3 = (0+0+0+1)%3 = 1


Now we have 2 numbers that are not divisible by 3 ?
Please correct me if i am wrong

[AlgorithmsMadeEasy]

If you haven't had fully understood how to solve the problem till now.
Check out my video on this question
Let me know your comments.

[MegaPaloma]

Please the audio is not good .. the sound is little too harsh for the ears .. but good video content keep going

[vatsalsonigara]

This is not O(n). The actual complexity is O(n*log(max_value))... because you have to check each bit O(n) times...and at max the bits are log(max_value)