@Atul

For array (3 2 5 2 5)
Sum+= (sum of number from 1 to n) , n is each array element.

Sum=(1+2+3)+(1+2)+(1+2+3+4+5)+(1+2)+(1+2+3+4+5) =42

On 1/4/12, SAMM <[email protected]> wrote:
> This may not work for the array having both +ve & -ve numbers becoz
> this logic is based on frequency distribution of elements from 1 to
> max(array elements) and if -ve num comes in here will disturb the
> frequency distribution.
>
> So this logic may work for only +ve and -ve elements array.
>
> On 1/4/12, SAMM <[email protected]> wrote:
>> I think this may works . needs verification.
>>
>> For the given array (3 5 2 5 2)
>> For +ve number (N) take the sum from 1 to N .
>> For -ve number (N) take the sum from -1 to N .
>> And take take the cumulative sum ... For this array it comes 42 .
>> Similarly check the sum for the second array . If it is same then we r
>> done
>> .
>>
>> On 1/3/12, atul anand <[email protected]> wrote:
>>> There are two arrays.
>>> int arr1[5] = { 3, 5, 2, 5, 2}
>>> int arr2[5] = { 2, 3, 5, 5, 2}
>>> The arrays will be called similar if they contain same number of
>>> elements
>>> equally.
>>> Write the pseudo code to check this ?
>>> not allowed to use sorting and hashtable.
>>>
>>> naive approach O(n^2)
>>>
>>> NOTE: Xoring , sum wont work.
>>>
>>> we can use O(n) space , using index as elements in the array. but if it
>>> has
>>> negative number then it will fail for eg arr1 has -1,...  and arr2 has
>>> 1,.....
>>>
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>>
>>
>> --
>> Somnath Singh
>>
>
>
> --
> Somnath Singh
>


-- 
Somnath Singh

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