Re: [algogeeks] Re: Valid Array

2010-12-10 Thread fenghuang
@mohan, when the num of repeatation is bigger than 1, it may be wrong,please
check {1, 1, 2, 5, 6, 6}

On Fri, Dec 10, 2010 at 12:41 PM, mo...@ismu mohan...@gmail.com wrote:

 i did nt get this xor part in adithya solution

 check if  this works

 array is valid if satisfy 2 conditions
 1.max-min=n-1
 2.there should be no repeatations

 first one can be done in O(n)
 for second

 check 1xor2xor...xorn=(a[1]-min+1)xor(a[2]-min+1)xor..
 if both are equal there are no repeatations

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[algogeeks] Re: Valid Array

2010-12-09 Thread Prims
I got the correct answer:

If it is a valid array, sum of all elements in the array = value
calculated using arithmetic progression formula

In this case

Sum of arithmetic progression = (n/2)[2*a+(n-1)*d}
where a = min of the array
n = number of elements
d = 1

If this value is equal to sum of all the elements in the array then it
is  a valid array

On Dec 10, 6:44 am, ADITYA KUMAR aditya...@gmail.com 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 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.- Hide quoted text -

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Re: [algogeeks] Re: Valid Array

2010-12-09 Thread mo...@ismu
prims  check for this case  [1,1,4,4]

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[algogeeks] Re: Valid Array

2010-12-09 Thread Prims
my bad

The solution i quoted works only in case of no repititions.

Aditya solution is correct

On Dec 10, 9:23 am, mo...@ismu mohan...@gmail.com wrote:
 prims  check for this case  [1,1,4,4]

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Re: [algogeeks] Re: Valid Array

2010-12-09 Thread mo...@ismu
i did nt get this xor part in adithya solution

check if  this works

array is valid if satisfy 2 conditions
1.max-min=n-1
2.there should be no repeatations

first one can be done in O(n)
for second

check 1xor2xor...xorn=(a[1]-min+1)xor(a[2]-min+1)xor..
if both are equal there are no repeatations

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