@rohit : why don't u have a look at the older posts before replying some
thing ...ok .. I'm sorry if u are hurt ..
what is the time complexity and space complexity of u'r older post ..

On Sat, Sep 10, 2011 at 8:03 AM, Rohit jalan <[email protected]> wrote:

> This can also be done:
>
>
> for( i=0; i<k;i++)
> {
>     b[i]=a[i];
> }
>
> for (;i<n;i++)
> {
>     a[i-k]=a[i];
> }
> while((i-k)<n)
> {
>     a[i-k]=b[i];
> }
>
> But extra array is used here.
>
>
> On Sat, Sep 10, 2011 at 5:30 PM, Rohit jalan <[email protected]> wrote:
>
>>
>> How is this one ??
>>
>>
>> int a[9]={9,7,6,5,3,23,14,2,4}
>> n=9
>> int *p,*q;
>> p=a;
>> for(q=a;i<n;q++,i++);
>> ReverseArray(p,q)
>>
>> p=&a[0];
>> q=&a[n-1-k]
>> ReverseArray(p,q)
>>
>> p=&a[n-k];
>> q=&a[n-1]
>> ReverseArray(p,q)
>>
>> void ReverseArray(int *l,int *r)
>> {
>>     int temp;
>>     while(l<r)
>>     {
>>         temp=*l;
>>         *l=*r;
>>         *r=temp
>>         l++;
>>         r--;
>>     }
>> }
>>
>>
>> Thanks & Regards,
>> -Rohit
>>
>>
>> On Sat, Sep 10, 2011 at 5:24 PM, bharatkumar bagana <
>> [email protected]> wrote:
>>
>>> swap k elements form 1 to k and n-k to n respectively...
>>> ex: k=3
>>>    temp=k;
>>>
>>> int a[9]= {9,7,6,5,3,23,14,2,4} ; has become {14,2,4,5,3,23,9,7,6};
>>>
>>>
>>>
>>> now swap first k elements with k+1 to 2k elements ...now k=2k+1 , do this 
>>> step again up to (k<n-temp)...
>>> at last {5,3,23,14,2,4,9,7,6,} ;
>>>
>>>
>>>
>>> Time :O(n) and space O(1).
>>>
>>>
>>>
>>> On Sat, Sep 10, 2011 at 7:15 AM, kumar raja <[email protected]>wrote:
>>>
>>>> @sarath:
>>>> I did not get u .Could u please explain it with the example.
>>>>
>>>>
>>>> On 10 September 2011 03:39, sarath prasath <[email protected]>wrote:
>>>>
>>>>> consider this approach..
>>>>> first reverse the entire array...
>>>>> so it will be.. 4,2,14,23,3,5,6,7,9
>>>>> and u want to shift k times right so
>>>>> u have to cut the array as n-k  and reverse both the sides u ll get
>>>>> it..
>>>>> so in ur scenario we are reversing upto the element 5 in array and
>>>>> reversing the remaining elements..
>>>>> hope the complexity is of o(n)..
>>>>>
>>>>>
>>>>>
>>>>> On Sat, Sep 10, 2011 at 3:17 PM, kumar raja 
>>>>> <[email protected]>wrote:
>>>>>
>>>>>> U have used c[3] extra array.It is already known solution. so it is
>>>>>> using O(k) space .i want the solution with constant space..
>>>>>>
>>>>>>
>>>>>> On 10 September 2011 02:08, Ishan Aggarwal <
>>>>>> [email protected]> wrote:
>>>>>>
>>>>>>> Solution :-
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> void main(){int a[9]= {9,7,6,5,3,23,14,2,4} ;int n = 3;int c[3];int 
>>>>>>> i;int k =0;for ( i=0;i<3;i++)
>>>>>>> c[i]= a[i];for(i=3;i<9;i++)
>>>>>>> a[i-3] =a[i];for(i=9-3;i<9;i++)
>>>>>>> a[i] = c[k++];for(i=0;i<9;i++)printf 
>>>>>>> <http://www.opengroup.org/onlinepubs/009695399/functions/printf.html>("\n%d",a[i]);}
>>>>>>>
>>>>>>>
>>>>>>> On Sat, Sep 10, 2011 at 2:09 PM, kumar raja <
>>>>>>> [email protected]> wrote:
>>>>>>>
>>>>>>>> Given an array of 'n' values you need to circular shift it 'k' times
>>>>>>>> towards left.
>>>>>>>>
>>>>>>>> Input : 9 7 6 5 3 23 14  2  4
>>>>>>>> output : 5 3 23 14 2  4 9 7  6
>>>>>>>>
>>>>>>>> n=9 , k= 3
>>>>>>>>
>>>>>>>> constraints : Time complexity O(n)
>>>>>>>> Space complexity O(1)
>>>>>>>>
>>>>>>>> The solutions with O(kn) time complexity and
>>>>>>>> O(n) complexity with O(k) space complexity are already available.
>>>>>>>>
>>>>>>>> I want the O(n) solution with constant space..
>>>>>>>> --
>>>>>>>> Regards
>>>>>>>> Kumar Raja
>>>>>>>> M.Tech(SIT)
>>>>>>>> IIT Kharagpur,
>>>>>>>> [email protected]
>>>>>>>> 7797137043.
>>>>>>>> 09491690115.
>>>>>>>>
>>>>>>>>  --
>>>>>>>> You received this message because you are subscribed to the Google
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>>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> --
>>>>>>> Kind Regards
>>>>>>> Ishan Aggarwal
>>>>>>> [image: Aricent Group]
>>>>>>> Presidency Tower-A, M.G.Road,Sector-14
>>>>>>> Gurgaon,Haryana.122015 INDIA
>>>>>>> Phone : +91-9654602663
>>>>>>> [email protected] <[email protected]>
>>>>>>>
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>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Regards
>>>>>> Kumar Raja
>>>>>> M.Tech(SIT)
>>>>>> IIT Kharagpur,
>>>>>> [email protected]
>>>>>> 7797137043.
>>>>>> 09491690115.
>>>>>>
>>>>>>  --
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>>>>>
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>>>>
>>>>
>>>>
>>>> --
>>>> Regards
>>>> Kumar Raja
>>>> M.Tech(SIT)
>>>> IIT Kharagpur,
>>>> [email protected]
>>>> 7797137043.
>>>> 09491690115.
>>>>
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>>>
>>>
>>>
>>> --
>>>
>>> **Please do not print this e-mail until urgent requirement. Go Green!!
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>>
>>
>>
>> --
>> Regards :
>> ROHIT JALAN
>> B.E. Graduate,
>> Computer Science Department,
>> RVCE, Bangalore
>>
>
>
>
> --
> Regards :
> ROHIT JALAN
> B.E. Graduate,
> Computer Science Department,
> RVCE, Bangalore
>
> --
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