its similar to performing a right rotatate within the specifie boundary the
number of times specified by argument
On Thu, Sep 16, 2010 at 11:49 PM, Aurelian Tutuianu <padre...@gmail.com>wrote:
> A naive way to implement that is:
>
> int[] v = new int[100]; // this is my vector with values
>
> for (int i = v.length - 1; i >= 0; i++) {
> // get maximum value
> int max = Integer.MIN_VALUE;
> for (int j = 0; j < v.length; j++) {
> if (max < v[j]) {
> max = j;
> }
> }
> // put max value on first position
> rev(max);
> // put max value on proper position
> rev(i);
> }
>
> 2*n calls to rev. (the fact that the find max is a naive approach does
> not count, according to problem)
>
> On Sep 16, 6:41 pm, Srinivas <lavudyasrinivas0...@gmail.com> wrote:
> > You have been given a function rev(x) which works as follows:
> > It reverses a[0] to a[x] elements of a.
> > e.g. Given array is
> > a : 1 8 3 4 5
> > rev(3) will convert the array to
> > a : 4 3 8 1 5
> > Use this function only (and comparison, of course) to sort given array
> > 'a'. The only criterion is
> > that the number of times this function is called should be minimum.
>
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