## 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:

**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.**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:

- Initialize two pointers:
`slowPointer`

and`fastPointer`

. Both pointers start at index 0. - 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`

.

- Repeat step 2 until the
`fastPointer`

reaches the end of the array. - 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];
slowPointer++;
}
}
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]
}
}
```

## Explanation

Let’s go through the solution step by step:

- We initialize
`slowPointer`

to 0, as it will be used to keep track of the modified array’s length. - We loop through the array using the
`fastPointer`

from index 0 to the last index. - 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. - If the element at
`fastPointer`

is equal to`val`

, we skip it and only increment the`fastPointer`

. - 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. - 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

**Empty Array**: If the input array is empty, the solution will correctly return a new length of 0.**All Elements are Equal to**: If all elements in the array are equal to the given value`val`

`val`

, the solution will still work. The`slowPointer`

will not move, and the original array will be effectively replaced with an empty subarray.

## Conclusion

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.