Write an algorithm that takes a sorted list of n integers and remove the duplicate elements from the list and return the new length. The algorithm must run in O (n) time and O (1) space. We assume that: List elements are inetger Input list is already sorted

integer = [1, 2, 4, 6, 7, 9, 9]

integer. sort ()

n = len (integer)

if n = = 0 or n = = 1:

print (n)

b = 0

for i in range (0, n-1):

if integer[i]! = integer[i+1]:

integer[b] = integer[i]

b + = 1

integer[b] = integer[n-1]

b + = 1

print (b)

Explanation:

The code is written in python

integer = [1, 2, 4, 6, 7, 9, 9]

The list that takes the list of integers

integer. sort ()

The list integer is sorted.

n = len (integer)

The length of the list of integer is gotten.

if n = = 0 or n = = 1:

If the length is equal to 0 or 1

print (n)

The the length

b = 0

We equate b to 0

for i in range (0, n-1):

The "for loop" loops through 0 to length of the list - 1

if integer[i]! = integer[i+1]:

if the index value in the list is not equals to the next index value in the list

integer[b] = integer[i]

The index value b is equal to index value of the looped item.

b + = 1

then add 1 to b

integer[b] = integer[n-1]

The index value of b = index value of the length - 1.

b + = 1

then add 1 to b

print (b)

print the value of b which is the length of the new list.