$*/$

public static Circle$2$DBig$[]$ sortCirclesByPerimeter$($Circle$2$DBig$[]$ ar$)$

$\{$

$//$write code

return null;

$\}$

$\frac{?}{?}****$

Sort the array in ascending order by the times a circle overlaps all

other circles in the array. It returns a new sorted array. You compare

every circle with all other circles. If overlaps with none should be

placed first. If overlaps with all others it should be placed last. Tie

is resolved by using perimeter$-$size. Smaller perimeter first.

@param ar the array to used for sorting, not altered.

@return a new new sorted array;

$*/$

public static Circle$2$DBig$[]$ sortCirclesByNumberOfTimesOverlapping$($Circle$2$DBig$[]$ ar$)$

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$//$write code

$\}$

return null;

$/**$

Prints the array of circles

@param ar the circles.

$*/$

public static void print$($Circle$2$DBig$[]$ ar$)$

$\{$

$//$write code

$\}$

@Override

public boolean equals$($Object obj$)$

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$//$replace this, write code

$\}$

return super.equals$($obj$)$;

@Override

public String toString$()$

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