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# Remove duplicates from an array of small primes

• Difficulty Level : Easy
• Last Updated : 15 Jun, 2021

Given an array of primes such that the range of primes is small. Remove duplicates from the array.

Examples:

```Input: arr[] = {3, 5, 7, 2, 2, 5, 7, 7};
Output: arr[] = {2, 3, 5, 7}
All the duplicates are removed from
the array. The output can be printed in any order.

Input: arr[] = {3, 5, 7, 3, 3, 13, 5, 13, 29, 13};
Output: arr[] = {3, 5, 7, 13, 29}
All the duplicates are removed from
the array. The output can be printed in any order.```

Source: Amazon Interview Question

Method 1: This method discusses the naive approach which takes O(n2) time complexity.

Approach: So the basic idea is to check for every element, whether it has occurred previously or not. Therefore, the approach involves keeping two loops, one to select the present element or index and the inner loop to check if the element has previously occurred or not.

Algorithm:

1. Start by running two loops.
2. Pick all elements one by one.
3. For every picked element, check if it has already occurred or not.
4. If already seen, then ignore it, else add it to the array.

## C++

 `// A C++ program to implement Naive``// approach to remove duplicates.``#include ``using` `namespace` `std;` `int` `removeDups(vector<``int``>& vect)``{``    ``int` `res_ind = 1;` `    ``// Loop invariant: Elements from vect``    ``// to vect[res_ind-1] are unique.``    ``for` `(``int` `i = 1; i < vect.size(); i++) {``        ``int` `j;``        ``for` `(j = 0; j < i; j++)``            ``if` `(vect[i] == vect[j])``                ``break``;``        ``if` `(j == i)``            ``vect[res_ind++] = vect[i];``    ``}` `    ``// Removes elements from vect[res_ind] to``    ``// vect[end]``    ``vect.erase(vect.begin() + res_ind, vect.end());``}` `// Driver code``int` `main()``{``    ``vector<``int``> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };``    ``removeDups(vect);``    ``for` `(``int` `i = 0; i < vect.size(); i++)``        ``cout << vect[i] << ``" "``;``    ``return` `0;``}`

## Java

 `// Java program to implement Naive``// approach to remove duplicates``class` `GFG {``    ``static` `int``[] removeDups(``int``[] vect)``    ``{``        ``int` `res_ind = ``1``;` `        ``// Loop invariant: Elements from vect``        ``// to vect[res_ind-1] are unique.``        ``for` `(``int` `i = ``1``; i < vect.length; i++) {``            ``int` `j;``            ``for` `(j = ``0``; j < i; j++)``                ``if` `(vect[i] == vect[j])``                    ``break``;``            ``if` `(j == i)``                ``vect[res_ind++] = vect[i];``        ``}` `        ``// Removes elements from vect[res_ind]``        ``// to vect[end]``        ``int``[] new_arr = ``new` `int``[res_ind];``        ``for` `(``int` `i = ``0``; i < res_ind; i++)``            ``new_arr[i] = vect[i];` `        ``return` `new_arr;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] vect = { ``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7` `};``        ``vect = removeDups(vect);` `        ``for` `(``int` `i = ``0``; i < vect.length; i++)``            ``System.out.print(vect[i] + ``" "``);``        ``System.out.println();``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# A Python3 program to implement``# Naive approach to remove duplicates.``def` `removeDups(vect):` `    ``res_ind ``=` `1` `# Loop invariant : Elements from vect``# to vect[res_ind-1] are unique.``    ``for` `i ``in` `range``(``1``, ``len``(vect)):``        ``j ``=` `0``        ``while` `(j < i):``            ``if` `(vect[i] ``=``=` `vect[j]):``                ``break``            ``j ``+``=` `1``        ``if` `(j ``=``=` `i):``            ``vect[res_ind] ``=` `vect[i]``            ``res_ind ``+``=` `1` `# Removes elements from``# vect[res_ind] to vect[end]``    ``return` `vect[``0``:res_ind]` `# Driver code``vect ``=` `[``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7``]``vect ``=` `removeDups(vect)``for` `i ``in` `range``(``len``(vect)):``    ``print``(vect[i], end ``=` `" "``)``    ` `# This code is contributed``# by mohit kumar`

## C#

 `// C# program to implement Naive approach``// to remove duplicates``using` `System;` `class` `GFG {``    ``static` `int``[] removeDups(``int``[] vect)``    ``{``        ``int` `res_ind = 1;` `        ``// Loop invariant : Elements from vect``        ``// to vect[res_ind-1] are unique.``        ``for` `(``int` `i = 1; i < vect.Length; i++) {``            ``int` `j;``            ``for` `(j = 0; j < i; j++)``                ``if` `(vect[i] == vect[j])``                    ``break``;``            ``if` `(j == i)``                ``vect[res_ind++] = vect[i];``        ``}` `        ``// Removes elements from vect[res_ind]``        ``// to vect[end]``        ``int``[] new_arr = ``new` `int``[res_ind];``        ``for` `(``int` `i = 0; i < res_ind; i++)``            ``new_arr[i] = vect[i];` `        ``return` `new_arr;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };``        ``vect = removeDups(vect);` `        ``for` `(``int` `i = 0; i < vect.Length; i++)``            ``Console.Write(vect[i] + ``" "``);``        ``Console.WriteLine();``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`3 5 7 2`

Complexity Analysis:

• Time Complexity: O(n2).
As two nested loops are used, so the time complexity becomes O(n2).
• Space Complexity: O(n).
As an extra array is used to store the elements, the space complexity is O(n).

Method 2: This method involves the technique of Sorting which takes O(n log n) time.

Approach: In comparison to the previous approach, a better solution is to first sort the array and then remove all the adjacent elements which are similar, from the sorted array.

Algorithm:

1. First sort the array.
2. The need for extra space can be cleverly avoided. Keeping two variables, first = 1 and i = 1.
3. Traverse the array from the second element to the end.
4. For every element, if that element is not equal to the previous element, then array[first++] = array[i], where i is the counter of a loop.
5. So the length of the array with no duplicates is first, remove the rest elements.

Note: In CPP there are a few inbuilt functions like sort() to sort and unique() to remove adjacent duplicates. The unique() function puts all unique elements at the beginning and returns an iterator pointing to the first element after the unique element. The erase() function removes elements between two given iterators.

## C++

 `// C++ program to remove duplicates using Sorting``#include ``using` `namespace` `std;` `int` `removeDups(vector<``int``>& vect)``{``    ``// Sort the vector``    ``sort(vect.begin(), vect.end());` `    ``// unique() removes adjacent duplicates.``    ``// unique function puts all unique elements at``    ``// the beginning and returns iterator pointing``    ``// to the first element after unique element.``    ``// Erase function removes elements between two``    ``// given iterators``    ``vect.erase(unique(vect.begin(), vect.end()),``               ``vect.end());``}` `// Driver code``int` `main()``{``    ``vector<``int``> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };``    ``removeDups(vect);``    ``for` `(``int` `i = 0; i < vect.size(); i++)``        ``cout << vect[i] << ``" "``;``    ``return` `0;``}`

## Java

 `// Java program to remove duplicates using Sorting``import` `java.util.*;` `class` `GFG {` `    ``static` `int``[] removeDups(``int``[] vect)``    ``{``        ``// sort the array``        ``Arrays.sort(vect);` `        ``// pointer``        ``int` `first = ``1``;` `        ``// remove duplicate elements``        ``for` `(``int` `i = ``1``; i < vect.length; i++)``            ``if` `(vect[i] != vect[i - ``1``])``                ``vect[first++] = vect[i];` `        ``// mark rest of elements to INT_MAX``        ``for` `(``int` `i = first; i < vect.length; i++)``            ``vect[i] = Integer.MAX_VALUE;` `        ``return` `vect;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] vect = { ``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7` `};``        ``vect = removeDups(vect);``        ``for` `(``int` `i = ``0``; i < vect.length; i++) {``            ``if` `(vect[i] == Integer.MAX_VALUE)``                ``break``;``            ``System.out.print(vect[i] + ``" "``);``        ``}``    ``}``}`

## Python3

 `# Python3 program to remove duplicates``# using Sorting``import` `sys` `def` `removeDups(vect):``    ` `    ``# Sort the array``    ``vect.sort()``    ` `    ``# Pointer``    ``first ``=` `1` `    ``# Remove duplicate elements``    ``for` `i ``in` `range``(``1``, ``len``(vect)):``        ``if` `(vect[i] !``=` `vect[i ``-` `1``]):``            ``vect[first] ``=` `vect[i]``            ``first ``+``=` `1``            ` `    ``# Mark rest of elements to INT_MAX``    ``for`  `i ``in` `range``(first, ``len``(vect)):``        ``vect[i] ``=` `sys.maxsize` `    ``return` `vect` `# Driver code``vect ``=` `[ ``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7` `]``vect ``=` `removeDups(vect)` `for` `i ``in` `range``(``len``(vect)):``    ``if` `(vect[i] ``=``=` `sys.maxsize):``        ``break``        ` `    ``print``(vect[i], end ``=` `" "``)` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# program to remove duplicates using Sorting``using` `System;``class` `GFG``{``    ``static` `int``[] removeDups(``int``[] vect)``    ``{``      ` `        ``// sort the array``        ``Array.Sort(vect);` `        ``// pointer``        ``int` `first = 1;` `        ``// remove duplicate elements``        ``for` `(``int` `i = 1; i < vect.Length; i++)``            ``if` `(vect[i] != vect[i - 1])``                ``vect[first++] = vect[i];` `        ``// mark rest of elements to INT_MAX``        ``for` `(``int` `i = first; i < vect.Length; i++)``            ``vect[i] = ``int``.MaxValue;``        ``return` `vect;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``params` `string``[] args)``    ``{``        ``int``[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };``        ``vect = removeDups(vect);``        ``for` `(``int` `i = 0; i < vect.Length; i++)``        ``{``            ``if` `(vect[i] == ``int``.MaxValue)``                ``break``;``            ``Console.Write(vect[i] + ``" "``);``        ``}``    ``}``}` `// This code is contributed by rutvik_56.`

## Javascript

 ``

Output:

`2 3 5 7`

Complexity Analysis:

• Time Complexity: O(n Log n).
For sorting the array O(n log n ) time complexity is required, and to remove adjacent elements O(n) time complexity is required.
• Auxiliary Space : O(1)
Since no extra space is required, the space complexity is constant.

Method 3: The method involves the technique of Hashing, which takes O(n) time.

Approach: The time complexity in this method can be reduced but space complexity will take a toll. This involves the use of Hashing where the numbers are marked in a HashMap, so that if the number is again encountered, then erase it from the array.

Algorithm:

1. Use a hash set. HashSet stores only unique elements.
2. It is known that if two of the same elements are put into a HashSet, the HashSet stores only one element (all the duplicate element vanishes)
3. Traverse the array from start to end.
4. For every element, insert the element into the HashSet
5. Now Traverse the HashSet and put the elements in the HashSet in the array

## C++

 `// C++ program to remove duplicates using Hashing``#include ``using` `namespace` `std;` `int` `removeDups(vector<``int``>& vect)``{``    ``// Create a set from vector elements``    ``unordered_set<``int``> s(vect.begin(), vect.end());` `    ``// Take elements from set and put back in``    ``// vect[]``    ``vect.assign(s.begin(), s.end());``}` `// Driver code``int` `main()``{``    ``vector<``int``> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };``    ``removeDups(vect);``    ``for` `(``int` `i = 0; i < vect.size(); i++)``        ``cout << vect[i] << ``" "``;``    ``return` `0;``}`

## Java

 `// Java program to implement Naive approach to``// remove duplicates.``import` `java.util.*;` `class` `GFG {` `    ``static` `void` `removeDups(Vector vect)``    ``{` `        ``// Create a set from vector elements``        ``Set set = ``new` `HashSet(vect);` `        ``// Take elements from set and put back in``        ``// vect[]``        ``vect.clear();``        ``vect.addAll(set);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``Integer arr[] = { ``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7` `};``        ``Vector vect``            ``= ``new` `Vector(``                ``Arrays.asList(arr));``        ``removeDups(vect);``        ``for` `(``int` `i = ``0``; i < vect.size(); i++) {``            ``System.out.print(vect.get(i) + ``" "``);``        ``}``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program to remove duplicates using Hashing``def` `removeDups():``    ``global` `vect` `    ``# Create a set from vector elements``    ``s ``=` `set``(vect)``    ` `    ``# Take elements from set and put back in``    ``# vect[]``    ``vect ``=` `s``    ` `# Driver code``vect ``=` `[``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7``]``removeDups()``for` `i ``in` `vect:``    ``print``(i, end ``=` `" "``)` `# This code is contributed by shubhamsingh10`

## C#

 `// C# program to implement Naive approach to``// remove duplicates.``using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `class` `GFG {` `    ``static` `List<``int``> removeDups(List<``int``> vect)``    ``{` `        ``// Create a set from vector elements``        ``HashSet<``int``> ``set` `= ``new` `HashSet<``int``>(vect);` `        ``// Take elements from set and put back in``        ``// vect[]``        ``vect.Clear();``        ``vect = ``set``.ToList();``        ``return` `vect;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 3, 5, 7, 2, 2, 5, 7, 7 };``        ``List<``int``> vect = ``new` `List<``int``>(arr);``        ``vect = removeDups(vect);``        ``for` `(``int` `i = 0; i < vect.Count; i++) {``            ``Console.Write(vect[i] + ``" "``);``        ``}``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`2 7 5 3`

Complexity Analysis:

• Time Complexity: O(n).
Since a single traversal is needed to enter all the elements in the hashmap, the time complexity is O(n).
• Auxiliary Space: O(n).
For storing elements in HashSet or hashmap O(n) space complexity is needed.

Method 4: This focuses on small ranged values where the time complexity is O(n).

Approach: This approach only works when the product of all distinct primes is fewer than 10^18 and all the numbers in the array should be prime. The property of primes of having no divisors except 1 or that number itself is used to arrive at the solution. As the array elements are removed from the array, they keep a value(product) which will contain the product of all distinct primes found previously in the array, so that if the element divides the product, they can be sure proved that the element has previously occurred in the array and hence the number will be rejected.

Algorithm:

1. Initially, keep a variable (p = 1).
2. Traverse the array from start to end.
3. While traversing, check whether p is divisible by the i-th element. If true, then erase that element.
4. Else keep that element and update the product by multiplying that element with the product (p = p * arr[i])

## C++

 `// Removes duplicates using multiplication of``// distinct primes in array``#include ``using` `namespace` `std;` `int` `removeDups(vector<``int``>& vect)``{``    ``long` `long` `int` `prod = vect;``    ``int` `res_ind = 1;``    ``for` `(``int` `i = 1; i < vect.size(); i++) {``        ``if` `(prod % vect[i] != 0) {``            ``vect[res_ind++] = vect[i];``            ``prod *= vect[i];``        ``}``    ``}``    ``vect.erase(vect.begin() + res_ind, vect.end());``}` `// Driver code``int` `main()``{``    ``vector<``int``> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };``    ``removeDups(vect);``    ``for` `(``int` `i = 0; i < vect.size(); i++)``        ``cout << vect[i] << ``" "``;``    ``return` `0;``}`

## Java

 `// Removes duplicates using multiplication of``// distinct primes in array``import` `java.util.*;` `class` `GFG {` `    ``static` `int``[] removeDups(``int``[] vect)``    ``{` `        ``int` `prod = vect[``0``];``        ``int` `res_ind = ``1``;``        ``for` `(``int` `i = ``1``; i < vect.length; i++) {``            ``if` `(prod % vect[i] != ``0``) {``                ``vect[res_ind++] = vect[i];``                ``prod *= vect[i];``            ``}``        ``}``        ``return` `Arrays.copyOf(vect, res_ind);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] vect = { ``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7` `};``        ``vect = removeDups(vect);``        ``for` `(``int` `i = ``0``; i < vect.length; i++)``            ``System.out.print(vect[i] + ``" "``);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Removes duplicates using multiplication of``# distinct primes in array``def` `removeDups(vect):``    ``prod ``=` `vect[``0``]``    ``res_ind ``=` `1``    ``i ``=` `1``    ``while` `(i < ``len``(vect)):``        ``if` `(prod ``%` `vect[i] !``=` `0``):``            ``vect[res_ind] ``=` `vect[i]``            ``res_ind ``+``=` `1``            ``prod ``*``=` `vect[i]``        ``vect ``=` `vect[:res_ind ``+` `2``]``        ``i ``+``=` `1``    ``return` `vect``    ` `# Driver code``vect ``=` `[``3``, ``5``, ``7``, ``2``, ``2``, ``5``, ``7``, ``7``]``vect ``=` `removeDups(vect)``for` `i ``in` `range``(``len``(vect)):``    ``print``(vect[i], end ``=``" "``)` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// Removes duplicates using multiplication of``// distinct primes in array``using` `System;` `class` `GFG {` `    ``static` `int``[] removeDups(``int``[] vect)``    ``{` `        ``int` `prod = vect;``        ``int` `res_ind = 1;``        ``for` `(``int` `i = 1; i < vect.Length; i++) {``            ``if` `(prod % vect[i] != 0) {``                ``vect[res_ind++] = vect[i];``                ``prod *= vect[i];``            ``}``        ``}``        ``int``[] temp = ``new` `int``[vect.Length - res_ind];``        ``Array.Copy(vect, 0, temp, 0, temp.Length);``        ``return` `temp;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };``        ``vect = removeDups(vect);``        ``for` `(``int` `i = 0; i < vect.Length; i++)``            ``Console.Write(vect[i] + ``" "``);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`3 5 7 2`

Complexity Analysis:

• Time Complexity: O(n).
To traverse the array only once, the time required is O(n).
• Auxiliary Space: O(1).
One variable p is needed, so the space complexity is constant.

Note: This solution would not work if there are any composites in the array.

This article is contributed by Shivam Mittal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.