🔹 Introduction

In programming, arrays often contain repeated elements. Removing duplicates ensures that only unique elements remain. This is essential in real-world applications like data cleaning, search optimization, and improving performance.

For example

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

There are several ways to remove duplicates from an array, each with different time and space complexities.

🛠️ Method 1. Using a Set (Recommended)

Most modern programming languages provide a Set data structure that automatically stores unique elements.

🔹 Java Example

import java.util.*;

public class RemoveDuplicates {
    public static void main(String[] args) {
        int[] arr = {1, 2, 2, 3, 4, 4, 5};
        Set<Integer> set = new LinkedHashSet<>(); // preserves order
        for (int num : arr) {
            set.add(num);
        }
        System.out.println("Unique elements: " + set);
    }
}

✅ Output

Unique elements: [1, 2, 3, 4, 5]

🔹 Python Example

arr = [1, 2, 2, 3, 4, 4, 5]
unique_arr = list(set(arr))
print("Unique elements:", unique_arr)

Note: In Python, converting to a set may not preserve the original order, unless you use dict.fromkeys(arr) or OrderedDict.

🛠️ Method 2. Using Sorting

Another approach is to sort the array first and then remove consecutive duplicates.

🔹 Java Example

import java.util.Arrays;

public class RemoveDuplicatesSorted {
    public static void main(String[] args) {
        int[] arr = {1, 2, 2, 3, 4, 4, 5};
        Arrays.sort(arr);
        int j = 0; // index for unique elements
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] != arr[j]) {
                j++;
                arr[j] = arr[i];
            }
        }
        int[] result = Arrays.copyOf(arr, j + 1);
        System.out.println("Unique elements: " + Arrays.toString(result));
    }
}

✅ Output

Unique elements: [1, 2, 3, 4, 5]

Time Complexity: O(n log n) (due to sorting)
Space Complexity: O(1)

🛠️ Method 3. Using a Temporary Array (Brute Force)

This method works by checking each element before adding it to a new array.

🔹 Java Example

public class RemoveDuplicatesBruteForce {
    public static void main(String[] args) {
        int[] arr = {1, 2, 2, 3, 4, 4, 5};
        int n = arr.length;
        int[] temp = new int[n];
        int j = 0;
        for (int i = 0; i < n; i++) {
            boolean isDuplicate = false;
            for (int k = 0; k < j; k++) {
                if (arr[i] == temp[k]) {
                    isDuplicate = true;
                    break;
                }
            }
            if (!isDuplicate) {
                temp[j++] = arr[i];
            }
        }
        int[] result = new int[j];
        System.arraycopy(temp, 0, result, 0, j);
        System.out.println("Unique elements: " + java.util.Arrays.toString(result));
    }
}

Time Complexity: O(n²)
Space Complexity: O(n)

This method is less efficient but useful to understand the basic logic behind removing duplicates.

💡 Tips for Removing Duplicates

🏆 Conclusion

Removing duplicates is a fundamental DSA problem that helps improve your coding and problem-solving skills. Understanding multiple approaches — Set, Sorting, Brute Force — prepares you for interviews and real-world projects.

By practicing this problem, you also learn about time-space tradeoffs, which is a key concept in efficient programming.