Remove Duplicates in-place from Sorted Array - Arrays

Problem Statement: Given an integer array sorted in non-decreasing order, remove the duplicates in place such that each unique element appears only once. The relative order of the elements should be kept the same.

If there are k elements after removing the duplicates, then the first k elements of the array should hold the final result. It does not matter what you leave beyond the first k elements.

Note: Return k after placing the final result in the first k slots of the array.

Examples:

Example 1: 

Input: arr[1,1,2,2,2,3,3]

Output: arr[1,2,3,_,_,_,_]

Explanation: Total number of unique elements are 3, i.e[1,2,3] and Therefore return 3 after assigning [1,2,3] in the beginning of the array.

Example 2: 

Input: arr[1,1,1,2,2,3,3,3,3,4,4]

Output: arr[1,2,3,4,_,_,_,_,_,_,_]

Explanation: Total number of unique elements are 4, i.e[1,2,3,4] and Therefore return 4 after assigning [1,2,3,4] in the beginning of the array.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1: Brute Force

Intuition: We have to think of a data structure that does not store duplicate elements. So can we use a Hash set? Yes! We can. As we know HashSet only stores unique elements.

Approach:

Declare a HashSet.
Run a for loop from starting to the end.
Put every element of the array in the set.
Store size of the set in a variable K.
Now put all elements of the set in the array from the starting of the array.
Return K.

Code:

C++ Code

#include<bits/stdc++.h>

using namespace std;
int removeDuplicates(int arr[]) {
  set < int > set;
  for (int i = 0; i < 7; i++) {
    set.insert(arr[i]);
  }
  int k = set.size();
  int j = 0;
  for (int x: set) {
    arr[j++] = x;
  }
  return k;
}
int main() {
  int arr[] = {1,1,2,2,2,3,3};
  int k = removeDuplicates(arr);
  cout << "The array after removing duplicate elements is " << endl;
  for (int i = 0; i < k; i++) {
    cout << arr[i] << " ";
  }
}

Output: The array after removing duplicate elements is 1 2 3

Time complexity: O(n*log(n))+O(n)

Space Complexity: O(n)

Java Code

import java.util.*;
public class Main {
    public static void main(String[] args) {
        int arr[] = {1,1,2,2,2,3,3};
        int k = removeDuplicates(arr);
        System.out.println("The array after removing duplicate elements is ");
        for (int i = 0; i < k; i++) {
            System.out.print(arr[i] + " ");
        }
    }
    static int removeDuplicates(int[] arr) {
        HashSet < Integer > set = new HashSet < > ();
        for (int i = 0; i < arr.length; i++) {
            set.add(arr[i]);
        }
        int k = set.size();
        int j = 0;
        for (int x: set) {
            arr[j++] = x;
        }
        return k;
    }
}

Output: The array after removing duplicate elements is 1 2 3

Time complexity: O(n*log(n))+O(n)

Space Complexity: O(n)

Solution 2: Two pointers

Intuition: We can think of using two pointers ‘i’ and ‘j’, we move ‘j’ till we don’t get a number arr[j] which is different from arr[i]. As we got a unique number we will increase the i pointer and update its value by arr[j].

Approach:

Take a variable i as 0;
Use a for loop by using a variable ‘j’ from 1 to length of the array.
If arr[j] != arr[i], increase ‘i’ and update arr[i] == arr[j].
After completion of the loop return i+1, i.e size of the array of unique elements.

Code:

C++ Code

#include<bits/stdc++.h>

using namespace std;
int removeDuplicates(int arr[]) {
  int i = 0;
  for (int j = 1; j < 7; j++) {
    if (arr[i] != arr[j]) {
      i++;
      arr[i] = arr[j];
    }
  }
  return i + 1;
}
int main() {
  int arr[] = {1,1,2,2,2,3,3};
  int k = removeDuplicates(arr);
  cout << "The array after removing duplicate elements is " << endl;
  for (int i = 0; i < k; i++) {
    cout << arr[i] << " ";
  }
}

Output:

The array after removing duplicate elements is 1 2 3

Time complexity: O(n)

Space Complexity: O(1)

Java Code

import java.util.*;
public class Main {
    public static void main(String[] args) {
       int arr[] = {1,1,2,2,2,3,3};
        int k = removeDuplicates(arr);
        System.out.println("The array after removing duplicate elements is ");
        for (int i = 0; i < k; i++) {
            System.out.print(arr[i] + " ");
        }
    }
    static int removeDuplicates(int[] arr) {
        int i = 0;
        for (int j = 1; j < arr.length; j++) {
            if (arr[i] != arr[j]) {
                i++;
                arr[i] = arr[j];
            }
        }
        return i + 1;
    }
}

Output:

The array after removing duplicate elements is 1 2 3

Time complexity: O(n)

Space Complexity: O(1)

Special thanks to Prashant Sahu for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article