yes O(n) solution posible..
1. traverse the first array..
2. for(i=1;i<=n;i++)
{ X[a[i]] = i ;
}
3. print array X[];
On Sat, Jan 1, 2011 at 10:42 PM, Soumya Prasad Ukil
<[email protected]>wrote:
> Is the second array really required, assuming intersection of both A & B is
> equal to either A or B?
>
> How about X[a[i]] = i?
>
>
> On 27 December 2010 21:24, Anand <[email protected]> wrote:
>
>> I have a two arrays
>>
>> One is
>>
>> 2 5 1 6 4 3
>>
>> other is
>>
>> 1 2 3 4 5 6.
>>
>> I want to make an array X which gives the index of its element on other
>> arrays.
>>
>> Meaning X[1] = 3 1 is element of the second array and 3 is the index of
>> element 1 in first array.
>>
>> How shall we get array X in O(nlogn).
>>
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>
>
>
> --
> regards,
> soumya prasad ukil
>
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