Handling ' 0 's
Case1: array containing single zero.
Caae 2: array containing multiple zeros.
int numberofZeros =0, index =0 ;
before[0]=1
after[length-1]=1;
for (int i=1; i<length;i++)
{
if( !a[i] ) //encountered zero
{
numberofZeros++;
if(numberofZeros == 1) index =i; //Case1 handling
before[i] = before[i-1];
after[length-1-i] = after[length-i];
}
else if (numberofZeros<=1)
{
before[i]=a[i]*before[i-1];
after[length-1-i]=a[length-1-i]*after[length-i];
}
else break; //Case2 handling
}
for(int i=0;i<length; i++) // replace array contents as per the question.
{
if(numberofZeros == 1)
{
if(i ==index)
a[index] = before[index] * after[index];
else
a[i] = 0;
}
else if(numberofZeros>1)
{
a[i]=0;
}
else
{
a[i] = before[i] * after[i];
}
}
Thanks,
NagRaaj.
On Sun, Sep 19, 2010 at 8:24 PM, Ashish Goel <ashg...@gmail.com> wrote:
> this is google question
>
> take arrays before[] and after
>
> before[0]=1
> after[length-1]=1;
>
> for (int i=1; i<length;i++)
> {
> before[i]=a[i]*before[i-1];
> after[length-1-i]=a[length-1-i]*after[length-i];
> }
>
> now resuly for the asked index is after[index]*before[index]
>
>
> the idea here is that we should not be using division..
>
>
> additionally, you can improve it if any of the numbers is zero
>
>
>
>
>
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Sun, Sep 19, 2010 at 8:18 PM, bittu <shashank7andr...@gmail.com> wrote:
>
>>
>> Given an array of numbers, replace each number with the product of
>> all the numbers in the array except the number itself *without* using
>> division.
>>
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