## The Problem Statement

Given an integer array `nums`

and an integer `val`

, the task is to remove all occurrences of `val`

in `nums`

in-place. Additionally, we need to return the number of elements in `nums`

that are not equal to `val`

.

To summarize, here are the key requirements:

- Modify the array
`nums`

in-place so that it contains only the elements that are not equal to`val`

. - Maintain the order of the elements in
`nums`

. - Return the count of elements in
`nums`

after removal.

### The Sliding Window Approach

At first glance, this problem might seem complex, but it can be elegantly solved using a sliding window approach. In a sliding window problem, we typically define a window (in this case, a range within the array) and move it across the array, updating it based on certain conditions.

### Step-by-Step Solution

Let's break down the provided solution code and explain how it achieves the desired outcome:

```
class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
l = 0 # Initialize a pointer for the left boundary of the window
for r in range(len(nums)): # Iterate through the array using a pointer for the right boundary
if nums[r] != val: # If the current element is not equal to val
nums[l] = nums[r] # Move the current element to the left boundary
l += 1 # Increment the left boundary pointer
return l # Return the count of elements not equal to val
```

**Initialization**: We initialize`l`

as a pointer representing the left boundary of our sliding window.**Iterating through the Array**: We iterate through the array using a pointer`r`

representing the right boundary of our sliding window.**Condition Check**: For each element`nums[r]`

, if it's not equal to`val`

, we include it in our window.**Moving Elements**: We move the non-`val`

elements to the left boundary of the array by assigning them to`nums[l]`

.**Incrementing Pointer**: After moving an element, we increment`l`

, effectively expanding our window.**Returning Result**: Finally, we return`l`

, which represents the count of elements in`nums`

that are not equal to`val`

.

### Conclusion

Through the sliding window approach, we've efficiently solved the problem of removing specific elements from an array while maintaining order. This solution not only accomplishes the task effectively but also showcases the versatility and power of the sliding window technique in algorithmic problem-solving.

By understanding the problem, breaking down the solution steps, and recognizing the sliding window pattern, we've equipped ourselves with a valuable tool for tackling similar challenges in the future. Happy coding!