Struct tuple
{int s;int e;}
tuple findsubarray(Array input)
{
int i=0, j=input.length()-1;
while(input[i] < input[i+1])
i++;
if(i==j) return NULL // array already sorted
while(input[j-1] < input[j])
j--;
// now the subarrays from 0 to i is sorted and j to end is sorted.
int min = min(input(i+1,j-1));
int max = max(input(i+1,j-1));
while(input[i]>min && i>0)
i--;
while(input[j]<max && j<input.size()-1)
j++;
return (i,j);
}
On Thu, Nov 24, 2011 at 6:09 PM, vikas <[email protected]> wrote:
> char arr[] = {'a', 'b', 'e', 'f', 'd', 'g', 'h', 'i'};
>
> calculate the point where array is not sorted , in this case arr[4] =
> 'd'
> calulate k in array[5..n] such that k > 4 arr[k] < d , take the min =
> min{ arr[k] }
> same thing for max from reverse
> use quick /selection sort to identify the correct insertion indices of
> min/max, that will be answer.
> complexity O(n)
>
> On Nov 24, 2:06 pm, Ankuj Gupta <[email protected]> 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.
>
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>
--
Nitin Garg
"Personality can open doors, but only Character can keep them open"
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