@Rishabh Deep Singh
–
Well, for the bits where x has a 1, n can have anything, and for the places where x has a 0, n must have a 0 too. Does that answer your question?

@Rishabh Deep Singh
–
That's an interesting exercise. If you are stuck on this, I'd urge you to consider writing your code recursively. That might help. Or if you are not concerned about efficiency, you may loop over all the values and check if it satisfies the condition.

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## Comments

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TopNewestSir are you CSE IITKGP?

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I am a Mechanical Student At IIT Kharagpur.

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Okay! I am CSE at IIT Patna.

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@Agnishom Chattopadhyay @Chew-Seong Cheong

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I Found a Blog about it on Codeforces as SOS Dynamic Programming @Agnishom Chattopadhyay

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That's cool. Send me the link.

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Link

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By next number, do you mean keeping

`x`

fixed, you want to search for the next such`n`

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I have edited the Problem a little bit.

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`x`

has a 1,`n`

can have anything, and for the places where`x`

has a 0,`n`

must have a`0`

too. Does that answer your question?Log in to reply

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