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Introduction to the “Remove Element” Interview Question

The “Remove Element” interview question is a common coding problem often encountered during technical interviews. It assesses a candidate’s understanding of array manipulation, in-place algorithms, and two-pointer techniques. The problem typically involves removing all instances of a specific value from an array while maintaining the order of the remaining elements. In this tutorial, we will explore the problem statement, discuss the key concepts, and provide a detailed solution in Java.

Problem Statement

Given an array nums and a value val, the task is to remove all instances of val from the array in-place and return the new length of the modified array. It’s important to note that the order of the elements that are not equal to val should be maintained, and the remaining elements after removing val can be shifted to fill the empty spaces.

Here’s the method signature:

public int removeElement(int[] nums, int val)

Key Concepts

Before we dive into the solution, let’s understand some key concepts that will help us solve the problem more effectively:

  1. In-Place Algorithm: An in-place algorithm operates directly on the input data structure and doesn’t require additional memory for operations. In this problem, we are required to remove elements in-place without using extra space.
  2. Two-Pointer Technique: The two-pointer technique involves using two pointers to traverse through an array simultaneously. It is particularly useful for problems that involve manipulation or comparison of elements within an array.

Approach to Solving the Problem

To solve the “Remove Element” problem, we will use the two-pointer technique to traverse the array and remove elements equal to the given value val. The key idea is to have one pointer (slowPointer) to keep track of the modified array’s length and another pointer (fastPointer) to traverse through the original array. As we encounter elements not equal to val, we will place them at the position indicated by the slowPointer, effectively removing instances of val from the array.

Let’s break down the approach into steps:

  1. Initialize two pointers: slowPointer and fastPointer. Both pointers start at index 0.
  2. Traverse the array using the fastPointer:
  • If the element at fastPointer is not equal to val, replace the element at index slowPointer with the element at index fastPointer. Then, increment both slowPointer and fastPointer.
  • If the element at fastPointer is equal to val, only increment fastPointer.
  1. Repeat step 2 until the fastPointer reaches the end of the array.
  2. The value of slowPointer will represent the new length of the modified array.

Java Solution

Let’s implement the approach in Java:

public class RemoveElementSolution {
    public int removeElement(int[] nums, int val) {
        int slowPointer = 0; // Initialize slowPointer
        for (int fastPointer = 0; fastPointer < nums.length; fastPointer++) {
            if (nums[fastPointer] != val) {
                nums[slowPointer] = nums[fastPointer];
        return slowPointer; // New length of the modified array

    public static void main(String[] args) {
        RemoveElementSolution solution = new RemoveElementSolution();
        int[] nums = {3, 2, 2, 3, 4, 5, 3};
        int val = 3;
        int newLength = solution.removeElement(nums, val);
        System.out.println("New length: " + newLength); // Output: New length: 4
        System.out.println("Modified array: " + Arrays.toString(nums)); // Output: Modified array: [2, 2, 4, 5, 5, 5, 3]


Let’s go through the solution step by step:

  1. We initialize slowPointer to 0, as it will be used to keep track of the modified array’s length.
  2. We loop through the array using the fastPointer from index 0 to the last index.
  3. Inside the loop, we check if the element at fastPointer is not equal to the given value val. If they are not equal, it means this element should be included in the modified array. We place this element at the position indicated by slowPointer and then increment both pointers (slowPointer and fastPointer) to continue the traversal.
  4. If the element at fastPointer is equal to val, we skip it and only increment the fastPointer.
  5. After the loop completes, the value of slowPointer will represent the new length of the modified array, and the array up to that length will contain the elements with instances of val removed.
  6. The modified array is now a subarray of the original array with a length of slowPointer, and we can print both the new length and the modified array to verify the results.

Complexity Analysis

The time complexity of this solution is O(n), where n is the length of the input array nums. We traverse through the array once using the two-pointer technique, and each operation within the loop takes constant time.

The space complexity is O(1) since the solution operates in-place without using any additional data structures that scale with the input size.

Handling Edge Cases

  1. Empty Array: If the input array is empty, the solution will correctly return a new length of 0.
  2. All Elements are Equal to val: If all elements in the array are equal to the given value val, the solution will still work. The slowPointer will not move, and the original array will be effectively replaced with an empty subarray.


In this tutorial, we explored the “Remove Element” interview question, which involves removing all instances of a given value from an array in-place while maintaining the order of the remaining elements. We discussed key concepts such as in-place algorithms and the two-pointer technique. We then presented a step-by-step approach to solving the problem in Java, along with a detailed explanation of the code. Finally, we analyzed the time and space complexity of the solution and considered how to handle edge cases.

Understanding this problem and its solution will not only help you excel in technical interviews but also enhance your problem-solving skills and algorithmic thinking in general.

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