that will report duplicate entries multiple times :(

On Mon, Jun 13, 2011 at 5:38 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:

> why do we need 2 bits at all ??
> i think single bit per table entry will do.
> say table is T[1025], and array is A[M]
>
>
> 1. Go through the N sized array and set bit 0 of the hash table entry to 1
> if it is present in the first array.
> 2. Go through the M sized array and if T[a[i]] is set then this element is
> there in second array else not.
>
> any cases this can fails??
>
> On Mon, Jun 13, 2011 at 5:28 PM, Divye Kapoor <divyekap...@gmail.com>wrote:
>
>> Use a hash table of size 1025 with bits per table entry = 2.
>> 1. Go through the N sized array and set bit 0 of the hash table entry to 1
>> if it is present in the first array.
>> 2. Go through the M sized array and set bit 1 of the hash table entry to 1
>> if the element belongs to 0 to 1024.
>> 3. Go through the hash table and print entries with bit values 11.
>>
>> Space usage = O(1), 2050 bits (rounded off to the corresponding nearest
>> byte). Time complexity = O(N + M)
>>
>> --
>> DK
>>
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>
>
>
> --
> Sunny Aggrawal
> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
>


-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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