*STEP-1*
Construct a self balancing Binary Search Tree where in the nodes represents
the elements of the given set..
*STEP-2*
Depending on the K (no. of partitions) keep a counter k
*STEP_3*
* * If BST not empty continue to Step-4 else exit with failure..
*STEP-4*
And while doing any traversal technique on BST keep on decrement the k
value and display the node values and delete them .
*STEP-5*
Once the first set is displayed again initialize k value and repeat Step-3
On Sun, Oct 30, 2011 at 6:41 PM, SAMMM wrote:
> An array is given consisting of real integers , can hav repeatative
> elements , Now u need to find K partitions whose union will hav all
> the elements of the array , but the intersection of any pair of
> cannot hav K elements in common.
>
> Exam:-
> Array- 1 1 2 2 3 1 4 3 5
>
> For K=3 the partition are :-
> (1 2 3) (1 5 2) (1 4 3)
>
>
> cann't be :- (1 2 3)(3 1 2)(1 4 5)
>
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