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 <[email protected]
<mailto:[email protected]>> 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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