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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