Re: [algogeeks] Valid Array

2010-12-12 Thread Terence

No. You made the same mistake as I.
Try this case: {1, 2, 2, 5, 5}.
Actually, this case defeats the solution of Manmeet's, yours, and mine.
(same min/max, same sum, same xor result)

I think the key point is that the N variable cannot be determined by 1 
or 2 equation constraint.



On 2010-12-10 9:44, ADITYA KUMAR wrote:

@jai
yeah, it can be done using count sort logic
but that will take O(n) extra space

which can be avoided by using XOR.

On Fri, Dec 10, 2010 at 3:34 AM, jai gupta sayhelloto...@gmail.com 
mailto:sayhelloto...@gmail.com wrote:


Algo:
In first traverse find the min and the max values.
if (max-min)  not equals (N-1)
return false
In next traverse map each in a hashtable of size N where
index=key-min. Now in case of collision return false
return true
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Aditya Kumar
B-tech 3rd year
Computer Science  Engg.
MNNIT, Allahabad.
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Re: [algogeeks] Valid Array

2010-12-09 Thread ADITYA KUMAR
O(2n) algo

1st traversal
calculate min
calculate sx=1^2^.^N
where N= no of elements

2nd traversal
fx=(A[1]-min+1)^(A[2]-min+1)...^(A[N]-min+1)

Now if sx=fx
Array is valid otherwise not


Array is called Valid if all the numbers appearing in A [1...N] are
 consecutive numbers.

 Example: A={5,3,4} is a valid array
 A={3,7,5,4,6} is a valid array

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Aditya Kumar
B-tech 3rd year
Computer Science  Engg.
MNNIT, Allahabad.

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Re: [algogeeks] Valid Array

2010-12-09 Thread jai gupta
Algo:
In first traverse find the min and the max values.
if (max-min)  not equals (N-1)
return false
In next traverse map each in a hashtable of size N where index=key-min. Now
in case of collision return false
return true

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Re: [algogeeks] Valid Array

2010-12-09 Thread ADITYA KUMAR
@jai
yeah, it can be done using count sort logic
but that will take O(n) extra space

which can be avoided by using XOR.

On Fri, Dec 10, 2010 at 3:34 AM, jai gupta sayhelloto...@gmail.com wrote:

 Algo:
 In first traverse find the min and the max values.
 if (max-min)  not equals (N-1)
 return false
 In next traverse map each in a hashtable of size N where index=key-min. Now
 in case of collision return false
 return true

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-- 
Regards
Aditya Kumar
B-tech 3rd year
Computer Science  Engg.
MNNIT, Allahabad.

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