Write an algorithm to delete the person at location Dp without destroying the ex

Question

Write an algorithm to delete the person at location Dp without destroying the existing order of the list. The front of the list (base) is located at location “F.” The current last item in the list at location “L,” and the last available location is at location “M.” Other locations in the list are utilized for other purposes and should not be touched.   You must detect and report underflow. On deletion, move all items above the deleted item down (towards F) to keep the list contiguous. Be sure to adjust L to properly reflect the number of people in the list. A sample deletion follows: I take off for “programs” written in any language rather than expressing the answer as an algorithm!

MyData: array(L0 .. M) of item;

MyData before of Joe :

Tom

Sam

Mary

Joe

Bob

1

2

3

4

5

6

7

8

L0

F

L

M

M+1

M+2

If the value of the pointer Dp is 4 (Sam) the resulting list will appear as shown below. Note the value in location 7 is physically in memory but not logically part of the list.

Tom

Mary

Joe

Bob

Bob

1

2

3

4

5

6

7

8

F

L

M

Tom

Sam

Mary

Joe

Bob

1

2

3

4

5

6

7

8

L0

F

L

M

M+1

M+2

Explanation / Answer

For Delete If you remove sam

Here F is First Position Considered tht will increase after delete dp SAM at 4th position

try this

if(F==0) then

underflow

else

item==array(F)

if(F==L)

set F=0

set R=0

else if (F==N) then

F=1

else

set front=front+1

print Array

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