[英語解説] LeetCode 27: Remove Element in-Place

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).

Brute Force Method

Time Complexity O(N^2)

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        k = len(nums)
        i = 0
        while i <= len(nums) - 1:
            if nums[i] == val and i < k:
                for j in range(i+1, len(nums)):
                    nums[j-1], nums[j] = nums[j], nums[j-1]
                k -= 1
                i -= 1 
            i += 1
        return k

Double Pointers (Fast and slow pointer)

The Two-Pointers Method, also referred to as the Fast-Slow Pointer Method, is an approach that enables the simultaneous execution of two loops within a single loop.

Here's the definition of each pointer:

Fast Pointer traverse the oringal array searching elements in the new array (which contains no element eqauls to val).

Slow Pointer indicates the position where the index of the new array is updated. It is the right boundary of the new target array.

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        slow = 0
        for fast in range(len(nums)):
            if nums[fast] != val:
                nums[fast], nums[slow] = nums[slow], nums[fast]
                slow += 1
        
        return slow

Double Pointers (Left & Right pointer)

we divide the array into two segments:

The first segment is the valid part, which stores elements that are not equal to "val."
The second segment is the invalid part, which stores elements that are equal to "val."

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
	left, right = 0, len(nums) - 1
        while left <= right:
            if nums[left] == val:
                nums[left], nums[right] = nums[right], nums[left]
                left -= 1
                right -= 1
            left += 1
        return left