Best approach to find duplicates in an array

The question ‘Find duplicate elements in an array‘ is frequently posed in interviews. Stay with us to verify ‘whether an array contains duplicates‘.

Welcome back, readers. In this blog post, I’ll be elucidating how to identify similar items within an array. During interviews, this concept may be presented as ‘Write an algorithm to detect duplicates in an array‘ or ‘Find duplicate pairs in an array‘. However, the approach remains consistent regardless of the phrasing.”

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Understand Question : Duplicate elements in array

If you were to consider the phrase ‘find duplicates in an array’, you might initially think it entails locating and displaying duplicate elements. However, this assumption would be incorrect. The task simply requires determining whether any elements are repeated within the array.

Is it confusing? Indeed, I understand why. If you encounter such ambiguity during runtime, be prepared to clarify during interviews where you might need to return either ‘true’ or ‘false’, or the duplicate elements themselves.

In this post, I will solely return a boolean value, either true or false.

Efficient way to find duplicates in array

In my case, during high school, I employed nested for loops to identify duplicates in an array. However, do you believe this approach is sufficient for acing an interview? I would argue not, as it lacks optimization. Therefore, stick with us to conquer the ‘Efficient way to find duplicates in an array’ interview question.

I’m planning to implement a HashSet to ascertain whether our array contains duplicates or not. This choice stems from the HashSet’s capability to search for any element with linear time complexity and its built-in functionality to check the uniqueness of elements. For further details about HashSet, you can refer to its official site.

Check Below Video Animation How this logic working. or Do check our Beginner and Intermediate pages

Program To Check An Array Contains Matching Elements

I intend to divide the solution for this program into two steps so that you can gain a deeper understanding of it.

Step 1

As is customary, in the first step, we’ll create the necessary variables to begin our logic. In the explanation of the logic, I’ve mentioned that I’m going to use a HashSet. Therefore, let’s create a variable called ‘list’.

private static bool IsContainsDuplicate(int[] data)
{
    HashSet<int> list = new HashSet<int>();   

    return false;
}

I’ve crafted a method that returns true or false based on whether a matching element is found in the array. It returns true if a matching element is found, otherwise it returns false.

Step 2

In this step, I’ll initiate a for loop to iterate through each element in the given list. I’ll check if the element is contained in the HashSet list. If it’s not present, I’ll add the element to the list using the add method. If the element is already present, I’ll return true.

private static bool IsContainsDuplicate(int[] data)
{
    HashSet<int> list = new HashSet<int>();   

    for (int i = 0; i < data.Length; i++)
    {
        if (list.Contains(data[i]))
        {
            return true;
        }
        else
        {
            list.Add(data[i]);
        }
    }

    return false;
}

If, after completing the iteration, no duplicate elements are found in the given array, we will return false.

Summarizing Check if array contains any duplicates

In this discussion, we explored the process of efficiently finding duplicate elements within an array, particularly focusing on interview scenarios. Initially, we acknowledged the common confusion surrounding the task and the need to clarify whether the goal is to return true/false or the duplicate elements themselves.

Moving forward, we highlighted the inefficiency of using nested for loops for this purpose, prompting the adoption of more optimized approaches. The suggestion was made to utilize a HashSet due to its ability to search for elements with linear time complexity and its built-in functionality for checking element uniqueness.

The solution was divided into two steps: first, the creation of necessary variables, such as the HashSet named ‘list’, and second, the implementation of a method to return true if duplicate elements are found and false otherwise.

In the method, a for loop iterated through each element in the array. If an element was not already present in the HashSet, it was added, and if it was already present, true was returned. Conversely, if no duplicate elements were found after completing the iteration, false was returned.