In this tutorial, we will dive into the “Remove Element” interview question, a commonly asked problem in coding interviews. We will explore the problem statement, discuss the approach to solving it, and provide a Python solution with step-by-step explanations. By the end of this tutorial, you will have a clear understanding of how to tackle this problem effectively.

## Problem Statement

**Problem:** Given an array `nums`

and a value `val`

, remove all instances of that value in-place and return the new length of the array. 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 order of elements can be changed. It doesn’t matter what you leave beyond the new length.

For example, if the input array is `[3, 2, 2, 3]`

and the value to remove is `3`

, the modified array could be `[2, 2]`

, and the function should return a length of `2`

.

## Approach

Before jumping into the solution, let’s discuss the approach we’ll take to solve this problem. Since the goal is to remove all instances of a given value from the array in-place, we can utilize two pointers technique: one for iterating through the array and another for keeping track of the next available position where a non-target value should be placed.

The basic idea is to iterate through the array, and whenever we encounter an element that is not equal to the target value, we place it at the position indicated by the second pointer. By doing this, we effectively overwrite the target value with non-target values, effectively removing them from the array.

## Python Solution

Let’s implement the approach discussed above into a Python solution. We’ll define a function called `remove_element`

that takes the array `nums`

and the value `val`

as parameters. The function will return the new length of the array after removing all instances of `val`

. Here’s the step-by-step solution:

```
def remove_element(nums, val):
# Initialize a pointer to keep track of the next available position
# for placing non-target values.
next_position = 0
# Iterate through the array using a pointer.
for num in nums:
if num != val:
# If the current element is not the target value,
# place it at the next available position and increment the pointer.
nums[next_position] = num
next_position += 1
# The value of 'next_position' indicates the new length of the modified array.
return next_position
```

## Walkthrough of the Solution

Let’s walk through the solution with an example to understand how it works. Consider the following input:

Input: `nums = [3, 2, 2, 3], val = 3`

- Initialize the
`next_position`

pointer to`0`

. - Start iterating through the array:

- For
`num = 3`

, which is equal to`val`

, do nothing. - For
`num = 2`

, which is not equal to`val`

, place it at`nums[0]`

(the first available position) and increment`next_position`

to`1`

. - For
`num = 2`

, again not equal to`val`

, place it at`nums[1]`

(the second available position) and increment`next_position`

to`2`

. - For
`num = 3`

, equal to`val`

, do nothing.

At the end of the iteration, the array `nums`

becomes `[2, 2, 2, 3]`

, and the value of `next_position`

is `2`

. This means the new length of the modified array is `2`

, as there are two non-target values present.

## Time and Space Complexity Analysis

The time complexity of this solution is O(n), where n is the length of the input array. We iterate through the array once, and each iteration involves constant time operations.

The space complexity is O(1) since we are modifying the input array in-place and not using any additional data structures that grow with the input size.

## Testing the Solution

To validate the correctness of our solution, let’s test it with a few examples:

```
# Test case 1
nums1 = [3, 2, 2, 3]
val1 = 3
print(remove_element(nums1, val1)) # Output: 2
print(nums1) # Output: [2, 2, 2, 3]
# Test case 2
nums2 = [0, 1, 2, 2, 3, 0, 4, 2]
val2 = 2
print(remove_element(nums2, val2)) # Output: 5
print(nums2) # Output: [0, 1, 3, 0, 4, 0, 4, 2]
```

## Conclusion

In this tutorial, we tackled the “Remove Element” interview question by using the two pointers technique to remove all instances of a given value from an array in-place. We discussed the problem statement, the approach to solving it, and provided a step-by-step Python solution. We also walked through an example and analyzed the time and space complexity of the solution. By understanding the concept of in-place modification and the two pointers technique, you are now well-equipped to solve similar array manipulation problems efficiently in coding interviews.