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 ?

On Sun, Dec 19, 2010 at 8:03 PM, Ankur Murarka
<ankur.murarka....@gmail.com> wrote:
> 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 <shoonya.mo...@gmail.com>
> 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]
>> 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]
>> --------------       --------------
>> sorted              sorted
>>
>>
>> now scan A[j] to A[k], and find any element that is smaller than any
>> number in A[0]-A[j], if any element is found, mark it as new j
>> similarly scan A[j]-A[k] and find any element that is larger than any
>> number in A[k]-A[n], if any element is found, mark it as new k
>>
>> final j and k are the answer...
>>
>>
>>
>> Mohit
>>
>>
>>
>> On Sun, Dec 19, 2010 at 2:32 AM, Dan <dant...@aol.com> wrote:
>>>
>>> On Dec 18, 9:57 am, snehal jain <learner....@gmail.com> 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.                :-)
>>>
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