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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