The problem ‘Remove Element’ is a frequent and important coding problem posed during interviews or coding challenges. It’s a notable problem on LeetCode, which evaluates your grasp of array manipulation and two-pointer technique in Python.

This article provides a comprehensive walkthrough of the problem, its solution, and the Python code necessary to execute this solution.

## Understanding the Problem Statement

The problem statement reads as follows:

**Given an array nums and a value val, remove all instances of that value in-place and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.**

The goal is to remove all occurrences of a specified value from an array. The key point is to do this “in-place”, that is, without allocating a new array or utilizing additional space to solve the problem. The expected output should be the length of the array after the element’s removal.

This problem is typically used to assess a candidate’s ability to manipulate arrays, particularly in-place operations, which are usually more efficient and less memory-consuming.

## Approach to the Problem

This problem can be solved using the two-pointer technique. The two-pointer technique is a handy method for solving array problems, particularly when the array is sorted or when we need to track two positions in the array.

One pointer (the slow runner, `i`

) keeps track of the current element in the array, while the other (the fast runner, `j`

) explores the rest of the array. We can iterate over the array with `j`

, if `nums[j]`

doesn’t equal to `val`

, we copy `nums[j]`

to `nums[i]`

and increment `i`

, if `nums[j]`

equals to `val`

, we skip this element by incrementing `j`

.

## Python Solution for Remove Element

Let’s implement this approach in Python:

```
class Solution:
def removeElement(self, nums, val):
i = 0
for j in range(len(nums)):
if nums[j] != val:
nums[i] = nums[j]
i += 1
return i
```

In this solution, we start with two pointers, `i`

and `j`

. We use a for loop to iterate over the array with `j`

. Inside the loop, if the element at index `j`

does not equal to `val`

, we replace `nums[i]`

with `nums[j]`

and increment `i`

. This is because we have found a valid element that we want to keep in the array.

After the loop, we return `i`

, which is the length of the array after removing the specified element.

This solution modifies the array `nums`

in place, thus satisfying the problem’s requirement. The space complexity is O(1), or constant space, and the time complexity is O(n), where n is the number of elements in the array.

## Conclusion

The ‘Remove Element’ problem is an insightful problem that can be solved efficiently using the two-pointer technique. It offers a good opportunity to practice in-place array manipulation and understand the balance between time and space complexity. Once you grasp the concept of maintaining two pointers – one for iterating over the array and another for keeping track of the valid elements, solving this problem becomes a straightforward task.