Remove duplicates from Sorted Array

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Given a sorted array arr[] of size n, the goal is to rearrange the array so that all distinct elements appear at the beginning in sorted order. Additionally, return the length of this distinct sorted subarray.

Note: The elements after the distinct ones can be in any order and hold any value, as they don’t affect the result.

Examples: 

Input: arr[] = [2, 2, 2, 2, 2]
Output: [2]
Explanation: All the elements are 2, So only keep one instance of 2.

Input: arr[] = [1, 2, 2, 3, 4, 4, 4, 5, 5]
Output: [1, 2, 3, 4, 5]

Input: arr[] = [1, 2, 3]
Output: [1, 2, 3]
Explanation : No change as all elements are distinct.

Using Hash Set – Works for Unsorted Also – O(n) Time and O(n) Space

• Use a hash set or dictionary to store elements already processed
• Initialize index of result array as 0.
• Traverse through the input array. If an element is not in the hash set, put it at the result index and insert into the set.

#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;

int removeDuplicates(vector<int>& arr) {
  
    // To track seen elements
    unordered_set<int> s; 
  
    // To maintain the new size of the array
    int idx = 0;  

    for (int i = 0; i < arr.size(); i++) {
        if (s.find(arr[i]) == s.end()) { 
            s.insert(arr[i]);  
            arr[idx++] = arr[i];  
        }
    }
 
    // Return the size of the array 
    // with unique elements
    return s.size(); 
}

int main() {
    vector<int> arr = {1, 2, 2, 3, 4, 4, 4, 5, 5};
    int newSize = removeDuplicates(arr);
    for (int i = 0; i < newSize; i++) 
        cout << arr[i] << " ";
    return 0;
}

import java.util.HashSet;

class GfG {

    static int removeDuplicates(int[] arr) {
        
        // To track seen elements
        HashSet<Integer> s = new HashSet<>();
        
        // To maintain the new size of the array
        int idx = 0;  

        for (int i = 0; i < arr.length; i++) {
            if (!s.contains(arr[i])) { 
                s.add(arr[i]);  
                arr[idx++] = arr[i];  
            }
        }

        // Return the size of the array 
        // with unique elements
        return idx;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 2, 3, 4, 4, 4, 5, 5};
        int newSize = removeDuplicates(arr);

        for (int i = 0; i < newSize; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}

def removeDuplicates(arr):
    
    # To track seen elements
    seen = set()
    
    # To maintain the new size of the array
    idx = 0

    for i in range(len(arr)):
        if arr[i] not in seen:
            seen.add(arr[i])
            arr[idx] = arr[i]
            idx += 1

    # Return the size of the array 
    # with unique elements
    return idx

if __name__ == "__main__":
    arr = [1, 2, 2, 3, 4, 4, 4, 5, 5]
    newSize = removeDuplicates(arr)

    for i in range(newSize):
        print(arr[i], end=" ")

using System;
using System.Collections.Generic;

class GfG {
    static int removeDuplicates(int[] arr) {
        
        // To track seen elements
        HashSet<int> s = new HashSet<int>();
        
        // To maintain the new size of the array
        int idx = 0;

        for (int i = 0; i < arr.Length; i++) {
            if (!s.Contains(arr[i])) { 
                s.Add(arr[i]);
                arr[idx++] = arr[i];
            }
        }

        // Return the size of the array 
        // with unique elements
        return idx;
    }

    static void Main() {
        int[] arr = {1, 2, 2, 3, 4, 4, 4, 5, 5};
        int newSize = removeDuplicates(arr);

        for (int i = 0; i < newSize; i++) {
            Console.Write(arr[i] + " ");
        }
    }
}

function removeDuplicates(arr) {
    
    // To track seen elements
    const s = new Set();
    
    // To maintain the new size of the array
    let idx = 0;

    for (let i = 0; i < arr.length; i++) {
        if (!s.has(arr[i])) {
            s.add(arr[i]);
            arr[idx++] = arr[i];
        }
    }

    // Return the size of the array 
    // with unique elements
    return idx;
}

// Driver code
const arr = [1, 2, 2, 3, 4, 4, 4, 5, 5];
const newSize = removeDuplicates(arr);

console.log(arr.slice(0, newSize).join(' '));

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Expected Approach – O(n) Time and O(1) Space

Since the array is sorted, we do not need to maintain a hash set. All occurrences of an element would be consecutive. So we mainly need to check if the current element is same as the previous element or not.

Step by step implementation:

• Start with idx = 1 (idx is going to hold the index of the next distinct item. Since there is nothing before the first item, we consider it as the first distinct item and begin idx with 1.
• Loop through the array for i = 0 to n-1.
  • At each index i, if arr[i] is different from arr[i-1], assign arr[idx] = arr[i] and increment idx.
• After the loop, arr[] contains the unique elements in the first idx positions.

#include <iostream>
#include <vector>
using namespace std;

int removeDuplicates(vector<int>& arr) {
    int n = arr.size();
    if (n <= 1)
        return n;

  	// Start from the second element
    int idx = 1; 
  
    for (int i = 1; i < n; i++) {
        if (arr[i] != arr[i - 1]) {
            arr[idx++] = arr[i];
        }
    }
    return idx;
}

int main() {
    vector<int> arr = {1, 2, 2, 3, 4, 4, 4, 5, 5};
    int newSize = removeDuplicates(arr);
    for (int i = 0; i < newSize; i++) 
        cout << arr[i] << " ";
    return 0;
}

#include <stdio.h>

int removeDuplicates(int arr[], int n) {
    if (n <= 1)
        return n;
	
  	// Start from the second element
    int idx = 1;
  
    for (int i = 1; i < n; i++) {
        if (arr[i] != arr[i - 1]) {
            arr[idx++] = arr[i];
        }
    }
    return idx;
}

int main() {
    int arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    int newSize = removeDuplicates(arr, n);
    
    for (int i = 0; i < newSize; i++) 
        printf("%d ", arr[i]);

    return 0;
}

class GfG {

    static int removeDuplicates(int[] arr) {
        int n = arr.length;
        if (n <= 1)
            return n;
		
      	// Start from the second element
        int idx = 1; 
      
        for (int i = 1; i < n; i++) {
            if (arr[i] != arr[i - 1]) {
                arr[idx++] = arr[i];
            }
        }
        return idx;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 2, 3, 4, 4, 4, 5, 5};
        int newSize = removeDuplicates(arr);

        for (int i = 0; i < newSize; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}

def removeDuplicates(arr):
    n = len(arr)
    if n <= 1:
        return n

    # Start from the second element
    idx = 1  
    
    for i in range(1, n):
        if arr[i] != arr[i - 1]:
            arr[idx] = arr[i]
            idx += 1

    return idx

if __name__ == "__main__":
    arr = [1, 2, 2, 3, 4, 4, 4, 5, 5]
    newSize = removeDuplicates(arr)

    for i in range(newSize):
        print(arr[i], end=" ")

using System;

class GfG {
    static int removeDuplicates(int[] arr) {
        int n = arr.Length;
        if (n <= 1)
            return n;
		
      	// Start from the second element
        int idx = 1; 
        for (int i = 1; i < n; i++) {
            if (arr[i] != arr[i - 1]) {
                arr[idx++] = arr[i];
            }
        }
        return idx;
    }

    static void Main() {
        int[] arr = {1, 2, 2, 3, 4, 4, 4, 5, 5};
        int newSize = removeDuplicates(arr);

        for (int i = 0; i < newSize; i++) {
            Console.Write(arr[i] + " ");
        }
    }
}

function removeDuplicates(arr) {
    const n = arr.length;
    if (n <= 1)
        return n;
	
    // Start from the second element
    let idx = 1; 
    for (let i = 1; i < n; i++) {
        if (arr[i] !== arr[i - 1]) {
            arr[idx++] = arr[i];
        }
    }

    return idx;
}

// Driver code
const arr = [1, 2, 2, 3, 4, 4, 4, 5, 5];
const newSize = removeDuplicates(arr);

console.log(arr.slice(0, newSize).join(' '));

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S

Sahil Chhabra

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