@Bhupendra
Thanks for pointing it out
actually, it should be
3.now first element in A[0]..A[j-1] greater than min is 7(index 1) and *LAST
* element in
A[k+1]A[n] less than max is 1(index 9)
if you look at code, it was proper
for( i = e+1; i < n; i++)
{
if(arr[i] < max)
e = i;
@Snehal -
Please avoid posting problems that have been solved somewhere else
already. That decreases the value of this group. More people will be
active here if we see new problems.
Hope you get the point, no offence intended.
-Swapnil
--
You received this message because you are subscribed to
simple and nice! n log n time...
I however believe there's a way to get the solution in O(n) time...
On Dec 19, 6:45 am, Ankur Khurana wrote:
> how about sorting the array in an auxillary array first. then compare
> the elements in sorted and un sorted array. the first elements to
> differ from
@mohit
suppose the input array is {4,7,8,6,5,11,13,17,0,1,2,3}
1.first step of Ur algorithm gives j=2,k=9 (index of 8 and 1)
2.in the second step min and max comes out to be 5 and 13
3.now first element in A[0]..A[j-1] greater than min is 7(index 1) and first
element in
A[k+1]A[n] less tha
i have posted this problem along with solution to my blog check :
http://code-forum.blogspot.com/2010/12/find-minimum-length-unsorted-subarray.html
On Dec 18, 10:57 pm, snehal jain wrote:
> Given an unsorted array arr[0..n-1] of size n, find the minimum length
> subarray arr[s..e] such that sorti
@Snehal
you can find example here
http://geeksforgeeks.org/?p=8858
hope this will clear things for all :)
Mohit
On Sun, Dec 19, 2010 at 8:42 PM, snehal jain wrote:
> @mohit
> can u plz explain ur algo with an example
>
> On Sun, Dec 19, 2010 at 8:15 PM, Ankur Khurana
> wrote:
>
>> how abou
@Ankur Murarka
didn't get you, how you got time complexity as exponential, it's linear
only.
Mohit
On Sun, Dec 19, 2010 at 8:15 PM, Ankur Khurana wrote:
> how about sorting the array in an auxillary array first. then compare
> the elements in sorted and un sorted array. the first elements to
@mohit
can u plz explain ur algo with an example
On Sun, Dec 19, 2010 at 8:15 PM, Ankur Khurana wrote:
> how about sorting the array in an auxillary array first. then compare
> the elements in sorted and un sorted array. the first elements to
> differ from start and end in unsorted array constit
how about sorting the array in an auxillary array first. then compare
the elements in sorted and un sorted array. the first elements to
differ from start and end in unsorted array constitute the sub array
we are finding. i dont know this solution seems to easy , it might be
wrong. Any comments ?
Doesnt the time complexity seem to be a li'l large?? Looks like its taking
exponential time...
On Sun, Dec 19, 2010 at 5:01 PM, mohit ranjan wrote:
>
> Let A[0..n] be the array
>
>
> Step 1: Start from A[0] and find out the first element, beyond which array
> in not sorted, let's call it A[j]
> S
Let A[0..n] be the array
Step 1: Start from A[0] and find out the first element, beyond which array
in not sorted, let's call it A[j]
Step 2: Start from A[n], move backward and find first element beyond which
array in not sorted, let's call it A[k]
so we have
A[0]A[j].A[k]A[n]
--
On Dec 18, 9:57 am, snehal jain wrote:
> Given an unsorted array arr[0..n-1] of size n, find the minimum length
> subarray arr[s..e] such that sorting this subarray makes the whole
> array sorted.
Sounds like a simple homework problem to me.:-)
--
You received this message beca
12 matches
Mail list logo
