@manish,
can you please tell other questions asked by ms today?
On Sun, Jul 17, 2011 at 4:53 PM, Piyush Sinha <ecstasy.piy...@gmail.com>wrote:
> It can be done using one extra array only that is the output array out[]
>
> *int out = (int *)malloc(sizeof(n)); //n is the number of elements in a
> int i,temp = 1;
>
> for(i=0;i<n;i++)
> {
> out[i] = temp;
> temp*=a[i];
> }
> temp =1;
> for(i=n-1;i>=0;i--)
> {
> out[i] *= temp;
> temp*=a[i];
> }*
>
>
>
> On Sun, Jul 17, 2011 at 4:43 PM, manish patel <manispatel...@gmail.com>wrote:
>
>> thanx!!!! the question was asked by MS today in MNNIT.
>>
>> On Sun, Jul 17, 2011 at 4:35 PM, Anurag Aggarwal <
>> anurag19aggar...@gmail.com> wrote:
>>
>>> take two extra arrays b[] and c[]
>>> in b[] store the following thing
>>> b[0]=1;
>>> b[i]=b[i-1]*a[i-1];
>>>
>>>
>>> in c[] store following things
>>> c[n-1]=1;
>>> c[i]=c[i+1]*a[i+1] (i>n-1)
>>> fill the c[] array in reverse order i.e. start from n-1 then go to 0;
>>>
>>> now output[] would be
>>> output[i]=b[i]*c[i];
>>>
>>>
>>>
>>>
>>> On Sun, Jul 17, 2011 at 4:28 PM, geek forgeek <geekhori...@gmail.com>wrote:
>>>
>>>> given an array a[0......n-1] .required to find the output array out
>>>> [0.........n-1] such that out [i] is the product of all the numbers a[0] to
>>>> a[n-1] excluding a[i]
>>>> for ex out[2]=a[0]*a[1]*a[3]*a[4]....a[n-1]
>>>> constraint is not using division operator
>>>>
>>>> how to do this in O(n)??
>>>>
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>>>
>>>
>>> Anurag Aggarwal
>>>
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>>
>>
>>
>> --
>> With Regards
>>
>> Manish Patel
>> BTech 2nd Year
>> Computer Science And Engineering
>> National Institute of Technology -Allahabad
>>
>>
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>
>
>
> --
> *Piyush Sinha*
> *IIIT, Allahabad*
> *+91-7483122727*
> * <https://www.facebook.com/profile.php?id=100000655377926> "NEVER SAY
> NEVER"
> *
>
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--
Best Regards,
Harshal Choudhary
7th Semester, CSE Dept.
NIT Surathkal, India.
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