ajaytank's blog

By ajaytank, 5 years ago, In English,

If n+1 integers a0,a1,a2.....an are given then how can we find efficiently the value of

nC0*a0 -nC1*a1 + nC2*a2 -nC3*a3 + ..... +/- nCn*an (mod 1000000007)

Actually this is the problem of hackerrank.com's SprintIndia qualification round.

n<=10^5 and time limit is 2 sec.

Thanks in advance.. :)

 
 
 
 
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5 years ago, # |
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Do you mean the value

modulo some prime? It's just straightforward summation, plus precalculated factorials and their modular inverses, in O(N).

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    5 years ago, # ^ |
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    Sorry, I forgot to mention to find the value mod 10^9+7.

    How can it be done in O(N)?

    For e.g n=100000, we need all these values 100000C0,100000C1,..... 100000C50000.... which is calculated in O(N*N)

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      5 years ago, # ^ |
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      plus precalculated factorials and their modular inverses

      If you don't know what a modular inverse is, look it up. Then, you can just use the formula to calculate a binomial coefficient in O(1) by multiplication.

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        5 years ago, # ^ |
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        k got it. thnks :)

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        5 years ago, # ^ |
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        If you use modular inverse, then you must calculate n! * ( k! * (n-k)! ) ^ (MOD-2) How can you do it in O(1) time.

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          5 years ago, # ^ |
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          Technically in , but I just considered that constant for simplicity. (Usually, it's better to pre-calculate modular inverses in case you need to use them more often, to save time.)

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            5 years ago, # ^ |
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            And you can calculate modular inverses to first n integers in time using this formula:

            So it is really

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              5 years ago, # ^ |
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              Could you explain your idea that calculating it in O(N) with details ?

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                5 years ago, # ^ |
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                Well, you want calculate modular inverses to first n factorials modulo p = 109 + 7.

                So you can calculate modular inverses to 1, 2, ..., n and get all factorial modular inverses in O(n) time.

                Modular inverses to 1, 2, ..., n can be calculated with dynamic programming (using above formula, note that , so if we now inverses to 1, 2, ..., k, we can get inverse to k + 1 in ).

                Proof of formula:

                Lets multiply both sides by

                We have:

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                  5 years ago, # ^ |
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                  Thank you for your explanation. I got it.