My bad, did not see the in-place memory requirement
On Sat, Dec 4, 2010 at 8:49 PM, coolfrog$
<dixit.coolfrog.div...@gmail.com>wrote:
> @shreyas VA
> you are using O(n) extra space...
>
>
> On Sat, Dec 4, 2010 at 8:46 PM, Shreyas VA <v.a.shre...@gmail.com> wrote:
>
>> Given the size of the input array,
>> construct array1 = {0, 1, 0, 1....} till n elements
>>
>> traverse through input array checking sum of 1's n 0's.
>>
>> at the end if both sums are equal return array1 else return input array.
>>
>> On Sat, Dec 4, 2010 at 12:06 AM, siva viknesh <sivavikne...@gmail.com>wrote:
>>
>>> Modified 2 color sort problem i.e. you are given an array of integers
>>> containing only 0s and 1s.You have to place all the 0s in even
>>> position and 1s in odd position. And if suppose, no. of 0s exceed no.
>>> of 1s or vice versa then keep them untouched. Do that in ONE PASS and
>>> without taking extra memory (modify the array in-place).
>>>
>>> For Example :
>>>
>>> Input Array: {0,1,1,0,1,0,1,0,1,1,1,0,0,1,0,1,1}
>>> Output Array: {0,1,0,1,0,1,0,1,0,1,0,1,0,1,1
>>> ,1,1}
>>>
>>> Write a solid secure code for it.
>>>
>>>
>>> My solution:
>>>
>>>
>>> .....i thought of a solution ..but it takes 2 passes !!
>>>
>>> in first pass count all no. of zeros nd ones
>>>
>>> and in 2nd pass check whether no. of zeros > no. of 1 s and vice
>>> versa and accordingly assign values to the same input array
>>>
>>> ....can anybody give the solution in single pass??
>>>
>>> ....
>>>
>>>
>>>
>>> --
>>> Regards,
>>> $iva
>>>
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>
>
>
> --
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