27.Remove_Element_(Easy).md

27. Remove Element (Easy)

Date and Time: Jun 4, 2024, 7:30 PM (EST)

Link: https://leetcode.com/problems/remove-element/

Question:

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.

• Return k.

Example 1:

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

Output: 2, nums = [2,2,,]

Explanation: Your function should return k = 2, with the first two elements of nums being 2. It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,1,2,2,3,0,4,2], val = 2

Output: 5, nums = [0, 1, 4, 0, 3,,,_]

Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4. Note that the five elements can be returned in any order. It does not matter what you leave beyond the returned k (hence they are underscores).

My Solution:

When there is no k == val, we perform in-place swap for every element in nums. When we see the element is val, we stop i at that position, and advance k to find the element that is not equal to val, then we replace ith position with that k value.

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        # Use a pointer to keep track of non-val index
        # Update pointer if current element in nums != val

        # TC: O(n), SC: O(1)
        i = 0
        for n in nums:
            if n != val:
                nums[i] = n
                i += 1
        return i

Time Complexity: $O(n)$
Space Complexity: $O(1)$