{
int odd=1,even=0;
while(odd <= size && even <=size)
{
if ( arr[even] != 1 )
even+=2;
if ( arr[odd] != 0 )
odd+=2;
if ( arr[even] == 1 && arr[odd] == 0 )
{
arr[even]=0;
arr[odd]=1;
}
}
}
On Thu, Aug 19, 2010 at 7:00 PM, ram das <ramnaraya...@gmail.com> wrote:
> shiftZeroOne(int *arr,int size)
>
> {
> int odd=1,even=0;
> while(odd <= size && even <=size)
> {
> if ( arr[even] != 1 )
> even+=2;
> if ( arr[odd] != 0 )
> odd+=2;
> if ( arr[even] == 1 && arr[odd] == 0 )
> {
> arr[even]=0;
> arr[odd]=1;
>
> }
> }
> }
>
>
>
>
> On Thu, Aug 19, 2010 at 5:53 PM, Rais Khan <raiskhan.i...@gmail.com>wrote:
>
>> void ArrayShifting(int *str, int size)
>> {
>> int odd=1, even=0;
>> while(odd < size)
>> {
>> int position;
>> if(str[even]!=0)
>> {
>> position = even+1;
>> while(position < size)
>> {
>> if(str[position]==0)
>> { str[position]=1;str[even]=0;break;}
>> position = position+2;
>> }
>> }
>> even = even+2;
>> if(str[odd]!=1)
>> {
>> position = odd+1;
>> while(position < size)
>> {
>> if(str[k]==0)
>> { str[position]=0;str[odd]=1;break;}
>> position = position+2;
>> }
>> }
>> odd=odd+2;
>> }
>> }
>>
>> This code seems working for me, If you agree then we can work on measuring
>> the complexity & improving the code accordingly.
>>
>>
>>
>>
>> On Thu, Aug 19, 2010 at 5:27 PM, Ashutosh Tamhankar <asshuto...@gmail.com
>> > wrote:
>>
>>> Hi Amit
>>>
>>> Am I allowed to keep counters to count the number of 1's and 0s right?
>>>
>>> The condition of not using extra memory is for the array!?
>>>
>>> Regards
>>> Ashutosh
>>>
>>>
>>> 2010/8/18 amit <amitjaspal...@gmail.com>
>>>
>>> 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 if 0s exceed no. of 1s or vice versa then keep them
>>>> untouched. Do that in ONE PASS and WITHOUT taking EXTRA MEMORY.
>>>>
>>>> --
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>>>>
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>>
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>
>
>
> --
> Thanks & Regards
> Ram Narayan Das
> mob: +91 9177711195
>
--
Thanks & Regards
Ram Narayan Das
mob: +91 9177711195
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