It is a well-researched fact that men in a restroom generally prefer to maximizetheir distance from already occupied stalls, by occupying the middle of the longestsequence of unoccupied places. for example, consider the situation where ten stalls are empty._ _ _ _ _ _ _ _ _ _the first visitor will occupy a middle position: _ _ _ _ _ x _ _ _ _the next visitor will be in the middle of the empty area at the left._ _ x _ _ x _ _ _ _write a program that reads the number of stalls and then prints out diagrams in theformat given above when the stalls become filled, one at a time. hint: use an array ofboolean values to indicate whether a stall is occupied.

The code below addresses the problem statement with appropriate comments for explanation.

Explanation:

public class Stall

{

public static void main(String[] args)

{

boolean[] arr = new boolean[10];

while(hasFreeSpots(arr))   // as long as there is a free slot

{

arr[position(arr)] = true;  

// position(arr) gets the position of slot a person would most likely enter at(view problem statement)

// setting that position true(position occupied) by arr[position(arr)]

System.out.println(stallView(arr));   // printing current state of the stall using 'X' and '_'

}

}

/**

* Find the empty position that is in the middle of the sub-array of largest group of empty spots

* @param arr an array of boolean values

* @return returns an an integer value. Optimal position to be occupied.

*/

public static int position(boolean[] arr)

{

int first = 0;   // first stores the starting position of "longest sequence of unoccupied places"

int last = 0;   // last stores the ending position of "longest sequence of unoccupied places"

int temp1 = 0;   // temp1 stores the starting position of "current sequence of unoccupied places"

int temp2 = 0; // temp2 stores the ending position of "current sequence of unoccupied places"

for(int i = 0; i < arr.length; i++)   // for each slot in the Stall

{    

if(!arr[i]) // if slot i is not occupied

{    

temp1 = i;   // store starting position of "current sequence of unoccupied places" in temp1

while(i < arr.length && !arr[i] )   // if i is a valid index and as long as this slot is not occupied increment i

{

i++;

}

// we break out of the loop either if i is invalid index(reached end of the array)

// or if the slot is occupied by a person already

temp2 = i;   // store ending position of "current sequence of unoccupied places" in temp2

if(Math.abs(first - last) < Math.abs(temp1 - temp2))

  // if current sequence is longer than longest sequence, update it

{

first = temp1;

last = temp2;    

}

}

}

return (first + last) / 2;   // return middle index of first and last

}

/**

* Checks if there are empty spots in an array, basically checks if there is a false boolean value

* thats representing an empty spot in an array

* @param arr the input array, of boolean values

* @return returns a boolean value, true if there are false elements inside the array

*/

public static boolean hasFreeSpots(boolean[] arr)

{

boolean rtn = true;   // initializing rtn to 'true'

for(int i = 0; i < arr.length; i++)   // for every slot in the Stall

{

if(!arr[i]){rtn = false;}   // sets rtn to false if a free spot found

}

return !rtn;   // returns true if a free spot found and returns false if no free spot found

}

/**

* X for true, _ for false. Representing occupied(X) and free stalls(_)

* @param arr a boolean array

* @return returns a string representation if the input

*/

public static String stallView(boolean[] arr)

{

String str = "";       // initializing str to be an empty String

for(int i = 0; i < arr.length; i++)   // for every slot in Stall

{

if(arr[i]){str += " X";}   // Add X to str if slot is occupied

else {str += " _";}           // prints _ to str if slot not occupied

}

return str;       // return str(pictorial representation of Stall with 'X', '_')

}

}

it would be a. excessive network collisions or d. slow connectivity between devices those would be my answer.

answerfc- file compare;