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10 November, 16:06

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

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  1. 10 November, 17:32
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    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.
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