@BharathKumar: extremely sorry dude .. will not do this again .. Can you
forgive me ? :p
On Sep 10, 2011 12:09 PM, "bharatkumar bagana" <[email protected]>
wrote:
> @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
>>>>>>>>> Groups "Algorithm Geeks" group.
>>>>>>>>> To post to this group, send email to [email protected].
>>>>>>>>> To unsubscribe from this group, send email to
>>>>>>>>> [email protected].
>>>>>>>>> For more options, visit this group at
>>>>>>>>> http://groups.google.com/group/algogeeks?hl=en.
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> --
>>>>>>>> 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]>
>>>>>>>>
>>>>>>>> --
>>>>>>>> You received this message because you are subscribed to the Google
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>>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> --
>>>>>>> Regards
>>>>>>> Kumar Raja
>>>>>>> M.Tech(SIT)
>>>>>>> IIT Kharagpur,
>>>>>>> [email protected]
>>>>>>> 7797137043.
>>>>>>> 09491690115.
>>>>>>>
>>>>>>> --
>>>>>>> You received this message because you are subscribed to the Google
>>>>>>> Groups "Algorithm Geeks" group.
>>>>>>> To post to this group, send email to [email protected].
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>>>>>>>
>>>>>>
>>>>>> --
>>>>>> You received this message because you are subscribed to the Google
>>>>>> Groups "Algorithm Geeks" group.
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>>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Regards
>>>>> Kumar Raja
>>>>> M.Tech(SIT)
>>>>> IIT Kharagpur,
>>>>> [email protected]
>>>>> 7797137043.
>>>>> 09491690115.
>>>>>
>>>>> --
>>>>> You received this message because you are subscribed to the Google
>>>>> Groups "Algorithm Geeks" group.
>>>>> To post to this group, send email to [email protected].
>>>>> To unsubscribe from this group, send email to
>>>>> [email protected].
>>>>> For more options, visit this group at
>>>>> http://groups.google.com/group/algogeeks?hl=en.
>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>>
>>>> **Please do not print this e-mail until urgent requirement. Go Green!!
>>>> Save Papers <=> Save Trees
>>>> *BharatKumar Bagana*
>>>> **http://www.google.com/profiles/bagana.bharatkumar<
http://www.google.com/profiles/bagana.bharatkumar>
>>>> *
>>>> Mobile +91 8056127652*
>>>> <[email protected]>
>>>>
>>>>
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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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>>
>
>
>
> --
>
> **Please do not print this e-mail until urgent requirement. Go Green!!
> Save Papers <=> Save Trees
> *BharatKumar Bagana*
> **http://www.google.com/profiles/bagana.bharatkumar<
http://www.google.com/profiles/bagana.bharatkumar>
> *
> Mobile +91 8056127652*
> <[email protected]>
>
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