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 i0)
i--;
while(input[j]max jinput.size()-1)
j++;
return (i,j);
}
On Thu, Nov 24, 2011 at 6:09 PM, vikas vikas.rastogi2...@gmail.com 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 ankuj2...@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.
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