Remove Element Leetcode Problem 27 [Python Solution]

In this blog post, we're going to tackle the LeetCode problem 27, Remove Element This problem falls under the category of Arrays & Hashing and is rated as "Easy." We will provide a Python solution to solve this problem efficiently.

Problem Overview

The problem statement is as follows: Given an integer array nums and an integer val, you need to remove all occurrences of val in the nums array in-place.

The order of the elements may change, but you need to return the number of elements in nums that are not equal to val.

This problem requires modifying the original array without allocating additional memory.

Understanding the Constraints

Before we dive into the solution, let's understand the constraints and the problem requirements:

• The length of the nums array can be between 0 and 100.
• The values in nums can range from 0 to 50.
• The value val can also range from 0 to 100.

Remove Element LeetCode Problem Solution

We will provide an efficient Python solution to solve this problem.

Efficient Approach:

Here's the Python code to efficiently solve the Remove Element problem:

def removeElement(nums, val):
    k = 0  # Initialize a pointer to keep track of non-val elements
    for i in range(len(nums)):
        if nums[i] != val:
            nums[k] = nums[i]
            k += 1
    return k  # Return the value of k, representing the count of non-val elements

This code snippet defines a function called removeElement that takes two arguments: nums (the input array) and val (the value to remove).

It uses two pointers, i and k, to efficiently process the array.

The core idea behind this approach is to iterate through the nums array, and when we encounter a non-val element, we place it at the position indicated by the k pointer.

This ensures that all non-val elements are moved to the beginning of the array.

The k pointer keeps track of the count of non-val elements, which will be the final length of the modified array.

Example:

Let's illustrate how this efficient approach works with an example:

Input:

nums = [3, 2, 2, 3]
val = 3


Output:

2 # This is the number of elements in the modified array that are not equal to 3


Modified nums array:

nums = [2, 2, _, _] # The underscores represent values that are not important


In this example, the efficient approach ensures that all non-val elements (in this case, the value 2) are placed at the beginning of the array, and the count of non-val elements is returned as the final length.

Time and Space Complexity

Let's analyze the time and space complexity of our efficient solution:

• Time Complexity: The algorithm makes a single pass through the nums array, which has a time complexity of O(n), where n is the length of the array.

• Space Complexity: The algorithm operates in-place and does not allocate any additional memory, resulting in a space complexity of O(1).

Reasoning Behind Our Approach

The reasoning behind this approach is based on efficiently processing the nums array in a single pass while ensuring that non-val elements are placed at the beginning of the array.

By using two pointers (i and k), we can avoid unnecessary value shuffling and achieve the desired result without extra space.

The key idea is to ignore val elements, copy non-val elements to their correct positions, and keep track of the count of non-val elements using the k pointer.

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Conclusion

In this blog post, we discussed the Remove Element problem, which involves removing all occurrences of a specified value from an integer array in-place.

We provided an efficient Python solution and explained the reasoning behind it.

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You can access the original problem statement on LeetCode.

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Happy coding!