How to Remove Duplicate Elements From Array in Javascript

Duplicates are always adjacent in a sorted array. For example, in the array {1,3,5,5,7,9}, 5 is a duplicate element. Let's write program to remove duplicate elements.

Method 1 – Using extra space

Suppose we want to remove duplicate elements from an array arr. For that, we will run a for loop in which i will point to the index of each element of arr. We will make another pointer j that will point to the index of the elements of a new array temp where non-duplicate elements will be stored. We will initialize j by 0 as temp is initially empty. Then we will check whether the i^th element of arr is equal to the (i+1)^th element .

 If it is, then

 -- We will increment i by 1

 else

 -- We will store the i^th element in temp. Then we will increment i by 1 and j by 1.

We will run the for loop from the 1st element till the second last element of arr. We will not include the last element in the for loop as there is no (i+1)^th element for the last element. After the loop is completed, store the last element in temp and increment j.

Then copy the j elements of temp to arr and print the same.

Example

Let us take an array arr[] ={1,3,5,5,7,9}. In the first iteration of the for loop, i will point to the index of the first element (1). Then we will check if the i^th element is equal to the (i+1)^th element.

1 is not equal to 3, so the i^th element (1) will be stored in another array temp and i will be incremented by 1 so that it points to the index of the second element (3).

Again, we will check whether 3 is equal to 5. It is not equal, so 3 will be stored in temp and i will be incremented by 1 so that it points to the index of the third element (5).

This time, the ith element (5) is equal to the (i+1)th element (5). So, the i^th element will not be stored in temp and i will be incremented so that it points to the index of the next 5.

Again, 5 is not equal to 7. So, 5 will be stored in temp and i will be incremented so that it points to the index of 7.

7 is not equal to 9. So, 7 will be stored in temp and i will be incremented to point to the index of 9.

At last, we will automatically add 9 to temp.

So, the array temp contains all non-duplicate elements of arr. Copy the j elements of temp to arr and print them.

Pseudo Code

                        j            =            0            // traverse elements of arr            for            i            =            0            to            n            -            2            // if ith element is not equal to (i+1)th element of arr, then store ith value in temp            if            arr            [            i            ]            !=            arr            [            i            +            1            ]            temp            [            j            ]            =            arr            [            i            ]            j            +=            1            // store last value of arr in temp            temp            [            j            ]            =            arr            [            n            -            1            ]            j            +=            1            // copy elements of temp to first j elements of arr            for            i            =            0            to            j            -            1            arr            [            i            ]            =            temp            [            i            ]            // print first j elements of array arr            for            i            =            0            to            j            -            1            print            arr            [            i            ]          

• C/C++
• Python
• Java

                        #include            <stdio.h>                        // function for removing duplicates            void            removeDuplicate            (            int            arr            [],            int            n            )            {            int            j            =            0            ;            int            temp            [            n            ];            // traverse elements of arr.                        for            (            int            i            =            0            ;            i            <            n            -            1            ;            i            ++            )            {            // if ith element is not equal to (i+1)th element of arr, then store ith value in temp            if            (            arr            [            i            ]            !=            arr            [            i            +            1            ])            {            temp            [            j            ]            =            arr            [            i            ];            j            ++            ;            }            }            // store last value of arr in temp            temp            [            j            ++            ]            =            arr            [            n            -            1            ];            // make array arr equal to temp            for            (            int            i            =            0            ;            i            <            j            ;            i            ++            )            {            arr            [            i            ]            =            temp            [            i            ];            }            // print final array arr            for            (            int            i            =            0            ;            i            <            j            ;            i            ++            )            {            printf            (            "%d "            ,            arr            [            i            ]);            }            }            int            main            ()            {            int            arr            []            =            {            1            ,            3            ,            5            ,            5            ,            7            ,            9            };            int            n            =            sizeof            (            arr            )            /            sizeof            (            arr            [            0            ]);            // calling function when number of elements in array is greater than 1            if            (            n            >            1            )            {            removeDuplicate            (            arr            ,            n            );            }            return            0            ;            }          

Output

1 3 5 7 9

Time Complexity

The time complexity is O(N).

Method 2 – Without using extra space

This method is the same as the previous one, but we will not make a seperate array (temp) like in the previous method. Instead, we will store our final resultant elements in the same array (arr) instead of temp.

Pseudo Code

                        j            =            0            // traverse elements of arr            for            i            =            0            to            n            -            2            // if ith element is not equal to (i+1)th element of arr, then store ith value in arr[j]            if            arr            [            i            ]            !=            arr            [            i            +            1            ]            arr            [            j            ]            =            arr            [            i            ]            j            +=            1            // store last value of arr in temp            arr            [            j            ]            =            arr            [            n            -            1            ]            j            +=            1            // print first j elements of array arr            for            i            =            0            to            j            -            1            print            arr            [            i            ]          

• C/C++
• Python
• Java

                        #include            <stdio.h>                        // function for removing duplicates            void            removeDuplicate            (            int            arr            [],            int            n            )            {            int            j            =            0            ;            // traverse elements of arr            for            (            int            i            =            0            ;            i            <            n            -            1            ;            i            ++            )            {            // if ith element is not equal to (i+1)th element, then store ith value in arr[j]            if            (            arr            [            i            ]            !=            arr            [            i            +            1            ])            {            arr            [            j            ]            =            arr            [            i            ];            j            ++            ;            }            }            // store last value of arr in arr[j]            arr            [            j            ++            ]            =            arr            [            n            -            1            ];            // print first j elements of array arr            for            (            int            i            =            0            ;            i            <            j            ;            i            ++            )            {            printf            (            "%d "            ,            arr            [            i            ]);            }            }            int            main            ()            {            int            arr            []            =            {            1            ,            3            ,            5            ,            5            ,            7            ,            9            };            int            n            =            sizeof            (            arr            )            /            sizeof            (            arr            [            0            ]);            // calling function when number of elements in array is greater than 1            if            (            n            >            1            )            {            removeDuplicate            (            arr            ,            n            );            }            return            0            ;            }          

Time Complexity

The time complexity is O(N).

How to Remove Duplicate Elements From Array in Javascript

Source: https://www.codesdope.com/blog/article/remove-duplicate-elements-from-sorted-array/

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