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@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.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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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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