https://github.com/ziadasem/remove-duplicated-elements

remove and check for duplicated elements in list with binary search
https://github.com/ziadasem/remove-duplicated-elements

Last synced: 4 months ago
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remove and check for duplicated elements in list with binary search

• Host: GitHub
• URL: https://github.com/ziadasem/remove-duplicated-elements
• Owner: ziadasem
• Created: 2022-06-22T23:56:12.000Z (over 3 years ago)
• Default Branch: main
• Last Pushed: 2022-06-23T00:08:09.000Z (over 3 years ago)
• Last Synced: 2025-08-21T01:25:14.695Z (5 months ago)
• Language: C++
• Size: 1.95 KB
• Stars: 1
• Watchers: 1
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: readme.md

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README

          
# Remove Duplicated Elements in a list

 

This algorithm detects duplicated elements in a list and removes them. it finds the duplicate number in a way similar to the binary search algorithm and iterates with the number of duplicated elements.

**limitations**

 - input list must be sorted

 - elements should have same step size i.e. 2,4,6,8 or 10,20,30,40

## working of code

 **1- Finding the number of duplicated elements**

      if the list is sorted and the difference between elements is equal then this equation should be true

    expected_list_length = last_element - first_element + 1

if this equation isn't true, the **difference between actual length and expected length** is the number of duplicates 

 **2- Finding the correct element**

       the state that the list should be is that every element should be equal to this condition 

 

    expected_element = first_element + difference * current_index

  

for example 

    [2,6,10,14,18], step = 4

 at index 3 element should be equal to **2 + 4*3 = 14**

 if an element is in its place -achieve the above equation- that means this element and the previous elements are in their position without duplication, so duplication will be in **next elements only**.

 

 **3- Finding the duplicated element**

       like binary search the middle element of the array is checked does it achieve the above the condition :

           **if true**: that means the left part is ordered and without duplication 

           then we will search in the right part.

           **if false**: that means the left part has a duplication so we will 

            update the index of middle of the list to be the **middle of the left part**

             and repeat the search until finding the duplicated element.

  **4- Remove all the duplicated elements**

       Repeat the previous steps with the number of the duplication 

 

## Testing

the following lists are inputs to the function

    {-3,-1,1,1,3},

	{-10,-7,-4,-1,2,2,5},

    {-1,0,0,0,0,0},

    {0,0,0,0,0,0},

	{-1,-1,-1,-1,-1,-1},

	

	{-3,-2,-1,0,0},

    {4,4,7,7,10,10,13,16},                              

    {1,2,3},

	{100,200,300,300,300,400,400,400},

    {11,14,17,20,23},

    

	{12,16,20,24,24},

    {11,11,17,23,23},

    {11,17,23,23,29},

    {11,17,23,29,29},

    {11,17,17,23,29},

the output

    -3, -1, 1, 3

	-10, -7, -4, -1, 2, 5

	-1, 0

	0

	-1

	

	-3, -2, -1, 0

	4, 7, 10, 13, 16

	1, 2, 3

	100, 200, 300, 400

	11, 14, 17, 20, 23

	

	12, 16, 20, 24

	11, 17, 23

	11, 17, 23, 29

	11, 17, 23, 29

	11, 17, 23, 29

## Help and support

if you have an improvement for this code or have failed test case, please share it with me