i think we need to maintain an array of index as well such that while
subtracting smallest element from largest element of sorted array we need to
check if index of largest is greater than index of smallest. if no..then
this is not the solution..
On 12 June 2010 14:20, Modeling Expert <cs.modelingexp...@gmail.com> wrote:
> Let's say array A , 1 till n
>
> s_index = 1; e_index = n ;
> start = &A[s_index] ;
> end = &A[e_index];
> result = 0; //! j - i
> if ( *end > *start ){
> result = index(end) - index(start) ( only of its greater than
> previous value of result )
> break ;
> }else{
> end-- ; //! till you reach start
> }
>
> now increment start , and repeat the process but only from A[n] till
> A[++result] , going further
> down is not required now.
>
> Average time < o(n^2)
>
> Worst case : let's say we have descending array of ints, theno(n^2)
>
> Please suggest improvements
>
>
>
>
>
>
>
>
>
>
> On Jun 12, 12:14 am, amit <amitjaspal...@gmail.com> wrote:
> > given an array A of n elements.
> > for indexes j , i such that j>i
> > maximize( j - i )
> > such that A[j] - A [ i ]> 0 .
> >
> > Any Algorithm less than O(n^2) would do.
>
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