In place solution to remove duplicates from a sorted list The Next CEO of Stack OverflowFinding longest common prefixRemove duplicates from a sorted arrayFind a number which equals to the total number of integers greater than itself in an arrayFirstDuplicate FinderGiven a sorted array nums, remove the duplicates in-placeRemove all occurrences of an element from an array, in placeRemove duplicates from sorted array, in placeHash table solution to twoSumA One-Pass Hash Table Solution to twoSummerge two sorted list in a decent solution but get low scores

How to start emacs in "nothing" mode (`fundamental-mode`)

If/When UK leaves the EU, can a future goverment conduct a referendum to join the EU?

Why am I allowed to create multiple unique pointers from a single object?

How to count occurrences of text in a file?

Are there any limitations on attacking while grappling?

Why do airplanes bank sharply to the right after air-to-air refueling?

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?

Should I tutor a student who I know has cheated on their homework?

What is the result of assigning to std::vector<T>::begin()?

Would a completely good Muggle be able to use a wand?

Can I run my washing machine drain line into a condensate pump so it drains better?

Anatomically Correct Strange Women In Ponds Distributing Swords

Written every which way

Does it take more energy to get to Venus or to Mars?

Which tube will fit a -(700 x 25c) wheel?

Sending manuscript to multiple publishers

Can I equip Skullclamp on a creature I am sacrificing?

How powerful is the invisibility granted by the Gloom Stalker ranger's Umbral Sight feature?

MessageLevel in QGIS3

Is it ever safe to open a suspicious html file (e.g. email attachment)?

Limits on contract work without pre-agreed price/contract (UK)

What connection does MS Office have to Netscape Navigator?

What does convergence in distribution "in the Gromov–Hausdorff" sense mean?

How did the Bene Gesserit know how to make a Kwisatz Haderach?



In place solution to remove duplicates from a sorted list



The Next CEO of Stack OverflowFinding longest common prefixRemove duplicates from a sorted arrayFind a number which equals to the total number of integers greater than itself in an arrayFirstDuplicate FinderGiven a sorted array nums, remove the duplicates in-placeRemove all occurrences of an element from an array, in placeRemove duplicates from sorted array, in placeHash table solution to twoSumA One-Pass Hash Table Solution to twoSummerge two sorted list in a decent solution but get low scores










6












$begingroup$


I am working on the problem removeDuplicatesFromSortedList




Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.



Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.



Example 1:



Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.



My solution and TestCase



class Solution: 
def removeDuplicates(self, nums: List[int]) -> int:
"""
"""
#Base Case
if len(nums) < 2: return len(nums)

#iteraton Case
i = 0 #slow-run pointer
for j in range(1, len(nums)):
if nums[j] == nums[i]:
continue
if nums[j] != nums[i]: #capture the result
i += 1
nums[i] = nums[j] #in place overriden
return i + 1

class MyCase(unittest.TestCase):
def setUp(self):
self.solution = Solution()

def test_raw1(self):
nums = [1, 1, 2]
check = self.solution.removeDuplicates(nums)
answer = 2
self.assertEqual(check, answer)

def test_raw2(self):
nums = [0,0,1,1,1,2,2,3,3,4]
check = self.solution.removeDuplicates(nums)
answer = 5
self.assertEqual(check, answer)

unittest.main()


This runs but I get a report:




Runtime: 72 ms, faster than 49.32% of Python3 online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 14.8 MB, less than 5.43% of Python3 online submissions for Remove Duplicates from Sorted Array.




Less than 5.43%, I employ the in-place strategies but get such a low rank, how could improve it?










share|improve this question









New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    This would be cheating, but did you try if the online judge actually checks if the array is manipulated in-place?
    $endgroup$
    – Graipher
    yesterday
















6












$begingroup$


I am working on the problem removeDuplicatesFromSortedList




Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.



Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.



Example 1:



Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.



My solution and TestCase



class Solution: 
def removeDuplicates(self, nums: List[int]) -> int:
"""
"""
#Base Case
if len(nums) < 2: return len(nums)

#iteraton Case
i = 0 #slow-run pointer
for j in range(1, len(nums)):
if nums[j] == nums[i]:
continue
if nums[j] != nums[i]: #capture the result
i += 1
nums[i] = nums[j] #in place overriden
return i + 1

class MyCase(unittest.TestCase):
def setUp(self):
self.solution = Solution()

def test_raw1(self):
nums = [1, 1, 2]
check = self.solution.removeDuplicates(nums)
answer = 2
self.assertEqual(check, answer)

def test_raw2(self):
nums = [0,0,1,1,1,2,2,3,3,4]
check = self.solution.removeDuplicates(nums)
answer = 5
self.assertEqual(check, answer)

unittest.main()


This runs but I get a report:




Runtime: 72 ms, faster than 49.32% of Python3 online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 14.8 MB, less than 5.43% of Python3 online submissions for Remove Duplicates from Sorted Array.




Less than 5.43%, I employ the in-place strategies but get such a low rank, how could improve it?










share|improve this question









New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    This would be cheating, but did you try if the online judge actually checks if the array is manipulated in-place?
    $endgroup$
    – Graipher
    yesterday














6












6








6





$begingroup$


I am working on the problem removeDuplicatesFromSortedList




Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.



Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.



Example 1:



Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.



My solution and TestCase



class Solution: 
def removeDuplicates(self, nums: List[int]) -> int:
"""
"""
#Base Case
if len(nums) < 2: return len(nums)

#iteraton Case
i = 0 #slow-run pointer
for j in range(1, len(nums)):
if nums[j] == nums[i]:
continue
if nums[j] != nums[i]: #capture the result
i += 1
nums[i] = nums[j] #in place overriden
return i + 1

class MyCase(unittest.TestCase):
def setUp(self):
self.solution = Solution()

def test_raw1(self):
nums = [1, 1, 2]
check = self.solution.removeDuplicates(nums)
answer = 2
self.assertEqual(check, answer)

def test_raw2(self):
nums = [0,0,1,1,1,2,2,3,3,4]
check = self.solution.removeDuplicates(nums)
answer = 5
self.assertEqual(check, answer)

unittest.main()


This runs but I get a report:




Runtime: 72 ms, faster than 49.32% of Python3 online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 14.8 MB, less than 5.43% of Python3 online submissions for Remove Duplicates from Sorted Array.




Less than 5.43%, I employ the in-place strategies but get such a low rank, how could improve it?










share|improve this question









New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am working on the problem removeDuplicatesFromSortedList




Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.



Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.



Example 1:



Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.



My solution and TestCase



class Solution: 
def removeDuplicates(self, nums: List[int]) -> int:
"""
"""
#Base Case
if len(nums) < 2: return len(nums)

#iteraton Case
i = 0 #slow-run pointer
for j in range(1, len(nums)):
if nums[j] == nums[i]:
continue
if nums[j] != nums[i]: #capture the result
i += 1
nums[i] = nums[j] #in place overriden
return i + 1

class MyCase(unittest.TestCase):
def setUp(self):
self.solution = Solution()

def test_raw1(self):
nums = [1, 1, 2]
check = self.solution.removeDuplicates(nums)
answer = 2
self.assertEqual(check, answer)

def test_raw2(self):
nums = [0,0,1,1,1,2,2,3,3,4]
check = self.solution.removeDuplicates(nums)
answer = 5
self.assertEqual(check, answer)

unittest.main()


This runs but I get a report:




Runtime: 72 ms, faster than 49.32% of Python3 online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 14.8 MB, less than 5.43% of Python3 online submissions for Remove Duplicates from Sorted Array.




Less than 5.43%, I employ the in-place strategies but get such a low rank, how could improve it?







python python-3.x programming-challenge memory-optimization






share|improve this question









New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited yesterday









Graipher

26.6k54092




26.6k54092






New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









AliceAlice

2164




2164




New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    This would be cheating, but did you try if the online judge actually checks if the array is manipulated in-place?
    $endgroup$
    – Graipher
    yesterday

















  • $begingroup$
    This would be cheating, but did you try if the online judge actually checks if the array is manipulated in-place?
    $endgroup$
    – Graipher
    yesterday
















$begingroup$
This would be cheating, but did you try if the online judge actually checks if the array is manipulated in-place?
$endgroup$
– Graipher
yesterday





$begingroup$
This would be cheating, but did you try if the online judge actually checks if the array is manipulated in-place?
$endgroup$
– Graipher
yesterday











2 Answers
2






active

oldest

votes


















8












$begingroup$

One way that might speed up your solution slightly is to only place the value if necessary. Also note that only one of the two if conditions can be true, so just use else. Or even better, since there was a continue in the other case, just don't indent it.



i = 0 #slow-run pointer
for j in range(1, len(nums)):
if nums[j] == nums[i]:
continue
# capture the result
i += 1
if i != j:
nums[i] = nums[j] #in place overriden


You can also save half of the index lookups by iterating over the values. Of course it will need slightly more memory this way.



def removeDuplicates(nums):
if len(nums) < 2:
return len(nums)
i = 0 #slow-run pointer
for j, value in enumerate(nums):
if value == nums[i]:
continue
# capture the result
i += 1
if i != j:
nums[i] = value # in place overriden
return i + 1


This can probably be further sped-up by saving nums[i] in a variable as well.




What is interesting is to see timing comparisons to using the itertools recipe unique_justseen (which I would recommend you use in production if you want to get duplicate free values in vanilla Python):



from itertools import groupby
from operator import itemgetter

def unique_justseen(iterable, key=None):
"List unique elements, preserving order. Remember only the element just seen."
# unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
# unique_justseen('ABBCcAD', str.lower) --> A B C A D
return map(next, map(itemgetter(1), groupby(iterable, key)))

def remove_duplicates(nums):
for i, value in enumerate(unique_justseen(nums)):
nums[i] = value
return i + 1


For nums = list(np.random.randint(100, size=10000)) they take almost the same time:




  • removeDuplicates took 0.0023s


  • remove_duplicates took 0.0026s





share|improve this answer











$endgroup$












  • $begingroup$
    @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
    $endgroup$
    – Graipher
    yesterday










  • $begingroup$
    @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
    $endgroup$
    – Graipher
    yesterday


















1












$begingroup$

A very minor concern: we have this condition:




 if len(nums) < 2: return len(nums)



but none of the included tests exercise it. If we want our testing to be complete, we should have cases with empty and 1-element lists as input.



TBH, I'd reduce that to a simpler condition, and remove the need for one of the tests:



if not nums:
return 0





share|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ifUsing("editor", function ()
    StackExchange.using("externalEditor", function ()
    StackExchange.using("snippets", function ()
    StackExchange.snippets.init();
    );
    );
    , "code-snippets");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "196"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );






    Alice is a new contributor. Be nice, and check out our Code of Conduct.









    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f216403%2fin-place-solution-to-remove-duplicates-from-a-sorted-list%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    One way that might speed up your solution slightly is to only place the value if necessary. Also note that only one of the two if conditions can be true, so just use else. Or even better, since there was a continue in the other case, just don't indent it.



    i = 0 #slow-run pointer
    for j in range(1, len(nums)):
    if nums[j] == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = nums[j] #in place overriden


    You can also save half of the index lookups by iterating over the values. Of course it will need slightly more memory this way.



    def removeDuplicates(nums):
    if len(nums) < 2:
    return len(nums)
    i = 0 #slow-run pointer
    for j, value in enumerate(nums):
    if value == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = value # in place overriden
    return i + 1


    This can probably be further sped-up by saving nums[i] in a variable as well.




    What is interesting is to see timing comparisons to using the itertools recipe unique_justseen (which I would recommend you use in production if you want to get duplicate free values in vanilla Python):



    from itertools import groupby
    from operator import itemgetter

    def unique_justseen(iterable, key=None):
    "List unique elements, preserving order. Remember only the element just seen."
    # unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
    # unique_justseen('ABBCcAD', str.lower) --> A B C A D
    return map(next, map(itemgetter(1), groupby(iterable, key)))

    def remove_duplicates(nums):
    for i, value in enumerate(unique_justseen(nums)):
    nums[i] = value
    return i + 1


    For nums = list(np.random.randint(100, size=10000)) they take almost the same time:




    • removeDuplicates took 0.0023s


    • remove_duplicates took 0.0026s





    share|improve this answer











    $endgroup$












    • $begingroup$
      @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
      $endgroup$
      – Graipher
      yesterday










    • $begingroup$
      @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
      $endgroup$
      – Graipher
      yesterday















    8












    $begingroup$

    One way that might speed up your solution slightly is to only place the value if necessary. Also note that only one of the two if conditions can be true, so just use else. Or even better, since there was a continue in the other case, just don't indent it.



    i = 0 #slow-run pointer
    for j in range(1, len(nums)):
    if nums[j] == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = nums[j] #in place overriden


    You can also save half of the index lookups by iterating over the values. Of course it will need slightly more memory this way.



    def removeDuplicates(nums):
    if len(nums) < 2:
    return len(nums)
    i = 0 #slow-run pointer
    for j, value in enumerate(nums):
    if value == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = value # in place overriden
    return i + 1


    This can probably be further sped-up by saving nums[i] in a variable as well.




    What is interesting is to see timing comparisons to using the itertools recipe unique_justseen (which I would recommend you use in production if you want to get duplicate free values in vanilla Python):



    from itertools import groupby
    from operator import itemgetter

    def unique_justseen(iterable, key=None):
    "List unique elements, preserving order. Remember only the element just seen."
    # unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
    # unique_justseen('ABBCcAD', str.lower) --> A B C A D
    return map(next, map(itemgetter(1), groupby(iterable, key)))

    def remove_duplicates(nums):
    for i, value in enumerate(unique_justseen(nums)):
    nums[i] = value
    return i + 1


    For nums = list(np.random.randint(100, size=10000)) they take almost the same time:




    • removeDuplicates took 0.0023s


    • remove_duplicates took 0.0026s





    share|improve this answer











    $endgroup$












    • $begingroup$
      @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
      $endgroup$
      – Graipher
      yesterday










    • $begingroup$
      @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
      $endgroup$
      – Graipher
      yesterday













    8












    8








    8





    $begingroup$

    One way that might speed up your solution slightly is to only place the value if necessary. Also note that only one of the two if conditions can be true, so just use else. Or even better, since there was a continue in the other case, just don't indent it.



    i = 0 #slow-run pointer
    for j in range(1, len(nums)):
    if nums[j] == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = nums[j] #in place overriden


    You can also save half of the index lookups by iterating over the values. Of course it will need slightly more memory this way.



    def removeDuplicates(nums):
    if len(nums) < 2:
    return len(nums)
    i = 0 #slow-run pointer
    for j, value in enumerate(nums):
    if value == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = value # in place overriden
    return i + 1


    This can probably be further sped-up by saving nums[i] in a variable as well.




    What is interesting is to see timing comparisons to using the itertools recipe unique_justseen (which I would recommend you use in production if you want to get duplicate free values in vanilla Python):



    from itertools import groupby
    from operator import itemgetter

    def unique_justseen(iterable, key=None):
    "List unique elements, preserving order. Remember only the element just seen."
    # unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
    # unique_justseen('ABBCcAD', str.lower) --> A B C A D
    return map(next, map(itemgetter(1), groupby(iterable, key)))

    def remove_duplicates(nums):
    for i, value in enumerate(unique_justseen(nums)):
    nums[i] = value
    return i + 1


    For nums = list(np.random.randint(100, size=10000)) they take almost the same time:




    • removeDuplicates took 0.0023s


    • remove_duplicates took 0.0026s





    share|improve this answer











    $endgroup$



    One way that might speed up your solution slightly is to only place the value if necessary. Also note that only one of the two if conditions can be true, so just use else. Or even better, since there was a continue in the other case, just don't indent it.



    i = 0 #slow-run pointer
    for j in range(1, len(nums)):
    if nums[j] == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = nums[j] #in place overriden


    You can also save half of the index lookups by iterating over the values. Of course it will need slightly more memory this way.



    def removeDuplicates(nums):
    if len(nums) < 2:
    return len(nums)
    i = 0 #slow-run pointer
    for j, value in enumerate(nums):
    if value == nums[i]:
    continue
    # capture the result
    i += 1
    if i != j:
    nums[i] = value # in place overriden
    return i + 1


    This can probably be further sped-up by saving nums[i] in a variable as well.




    What is interesting is to see timing comparisons to using the itertools recipe unique_justseen (which I would recommend you use in production if you want to get duplicate free values in vanilla Python):



    from itertools import groupby
    from operator import itemgetter

    def unique_justseen(iterable, key=None):
    "List unique elements, preserving order. Remember only the element just seen."
    # unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
    # unique_justseen('ABBCcAD', str.lower) --> A B C A D
    return map(next, map(itemgetter(1), groupby(iterable, key)))

    def remove_duplicates(nums):
    for i, value in enumerate(unique_justseen(nums)):
    nums[i] = value
    return i + 1


    For nums = list(np.random.randint(100, size=10000)) they take almost the same time:




    • removeDuplicates took 0.0023s


    • remove_duplicates took 0.0026s






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited yesterday

























    answered yesterday









    GraipherGraipher

    26.6k54092




    26.6k54092











    • $begingroup$
      @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
      $endgroup$
      – Graipher
      yesterday










    • $begingroup$
      @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
      $endgroup$
      – Graipher
      yesterday
















    • $begingroup$
      @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
      $endgroup$
      – Graipher
      yesterday










    • $begingroup$
      @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
      $endgroup$
      – Graipher
      yesterday















    $begingroup$
    @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
    $endgroup$
    – Graipher
    yesterday




    $begingroup$
    @Aethenosity I read it as both improvements would be welcome (and since they accepted the answer, apparently it was not so far off the mark either). Note that answers are free to comment on any and all aspects of the code here, anyways, regardless of what the OP wants.
    $endgroup$
    – Graipher
    yesterday












    $begingroup$
    @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
    $endgroup$
    – Graipher
    yesterday




    $begingroup$
    @Aethenosity Feel free to post another answer if you see a way to reduce the memory used, though!
    $endgroup$
    – Graipher
    yesterday













    1












    $begingroup$

    A very minor concern: we have this condition:




     if len(nums) < 2: return len(nums)



    but none of the included tests exercise it. If we want our testing to be complete, we should have cases with empty and 1-element lists as input.



    TBH, I'd reduce that to a simpler condition, and remove the need for one of the tests:



    if not nums:
    return 0





    share|improve this answer









    $endgroup$

















      1












      $begingroup$

      A very minor concern: we have this condition:




       if len(nums) < 2: return len(nums)



      but none of the included tests exercise it. If we want our testing to be complete, we should have cases with empty and 1-element lists as input.



      TBH, I'd reduce that to a simpler condition, and remove the need for one of the tests:



      if not nums:
      return 0





      share|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        A very minor concern: we have this condition:




         if len(nums) < 2: return len(nums)



        but none of the included tests exercise it. If we want our testing to be complete, we should have cases with empty and 1-element lists as input.



        TBH, I'd reduce that to a simpler condition, and remove the need for one of the tests:



        if not nums:
        return 0





        share|improve this answer









        $endgroup$



        A very minor concern: we have this condition:




         if len(nums) < 2: return len(nums)



        but none of the included tests exercise it. If we want our testing to be complete, we should have cases with empty and 1-element lists as input.



        TBH, I'd reduce that to a simpler condition, and remove the need for one of the tests:



        if not nums:
        return 0






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered yesterday









        Toby SpeightToby Speight

        26.9k742118




        26.9k742118




















            Alice is a new contributor. Be nice, and check out our Code of Conduct.









            draft saved

            draft discarded


















            Alice is a new contributor. Be nice, and check out our Code of Conduct.












            Alice is a new contributor. Be nice, and check out our Code of Conduct.











            Alice is a new contributor. Be nice, and check out our Code of Conduct.














            Thanks for contributing an answer to Code Review Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f216403%2fin-place-solution-to-remove-duplicates-from-a-sorted-list%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            រឿង រ៉ូមេអូ និង ហ្ស៊ុយលីយេ សង្ខេបរឿង តួអង្គ បញ្ជីណែនាំ

            Crop image to path created in TikZ? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Crop an inserted image?TikZ pictures does not appear in posterImage behind and beyond crop marks?Tikz picture as large as possible on A4 PageTransparency vs image compression dilemmaHow to crop background from image automatically?Image does not cropTikzexternal capturing crop marks when externalizing pgfplots?How to include image path that contains a dollar signCrop image with left size given

            Romeo and Juliet ContentsCharactersSynopsisSourcesDate and textThemes and motifsCriticism and interpretationLegacyScene by sceneSee alsoNotes and referencesSourcesExternal linksNavigation menu"Consumer Price Index (estimate) 1800–"10.2307/28710160037-3222287101610.1093/res/II.5.31910.2307/45967845967810.2307/2869925286992510.1525/jams.1982.35.3.03a00050"Dada Masilo: South African dancer who breaks the rules"10.1093/res/os-XV.57.1610.2307/28680942868094"Sweet Sorrow: Mann-Korman's Romeo and Juliet Closes Sept. 5 at MN's Ordway"the original10.2307/45957745957710.1017/CCOL0521570476.009"Ram Leela box office collections hit massive Rs 100 crore, pulverises prediction"Archived"Broadway Revival of Romeo and Juliet, Starring Orlando Bloom and Condola Rashad, Will Close Dec. 8"Archived10.1075/jhp.7.1.04hon"Wherefore art thou, Romeo? To make us laugh at Navy Pier"the original10.1093/gmo/9781561592630.article.O006772"Ram-leela Review Roundup: Critics Hail Film as Best Adaptation of Romeo and Juliet"Archived10.2307/31946310047-77293194631"Romeo and Juliet get Twitter treatment""Juliet's Nurse by Lois Leveen""Romeo and Juliet: Orlando Bloom's Broadway Debut Released in Theaters for Valentine's Day"Archived"Romeo and Juliet Has No Balcony"10.1093/gmo/9781561592630.article.O00778110.2307/2867423286742310.1076/enst.82.2.115.959510.1080/00138380601042675"A plague o' both your houses: error in GCSE exam paper forces apology""Juliet of the Five O'Clock Shadow, and Other Wonders"10.2307/33912430027-4321339124310.2307/28487440038-7134284874410.2307/29123140149-661129123144728341M"Weekender Guide: Shakespeare on The Drive""balcony"UK public library membership"romeo"UK public library membership10.1017/CCOL9780521844291"Post-Zionist Critique on Israel and the Palestinians Part III: Popular Culture"10.2307/25379071533-86140377-919X2537907"Capulets and Montagues: UK exam board admit mixing names up in Romeo and Juliet paper"Istoria Novellamente Ritrovata di Due Nobili Amanti2027/mdp.390150822329610820-750X"GCSE exam error: Board accidentally rewrites Shakespeare"10.2307/29176390149-66112917639"Exam board apologises after error in English GCSE paper which confused characters in Shakespeare's Romeo and Juliet""From Mariotto and Ganozza to Romeo and Guilietta: Metamorphoses of a Renaissance Tale"10.2307/37323537323510.2307/2867455286745510.2307/28678912867891"10 Questions for Taylor Swift"10.2307/28680922868092"Haymarket Theatre""The Zeffirelli Way: Revealing Talk by Florentine Director""Michael Smuin: 1938-2007 / Prolific dance director had showy career"The Life and Art of Edwin BoothRomeo and JulietRomeo and JulietRomeo and JulietRomeo and JulietEasy Read Romeo and JulietRomeo and Julieteeecb12003684p(data)4099369-3n8211610759dbe00d-a9e2-41a3-b2c1-977dd692899302814385X313670221313670221