RemoveElement

Problem

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

Solution

Start from the first element of nums. If the current element nums[i] equals val, look ahead to the next element by incrementing an offset. If the current element nums[i + offset] is unequal to val, then copy nums[i + offset] to nums[i] and increment i. Repeat until i + offset is out of bounds.

Code

from typing import List


def removeElement(nums: List[int], val: int) -> int:
    """Removes all instances of ``val`` from the array ``nums`` in-place.
    """
    i = 0
    offset = 0
    while i + offset < len(nums):
        if nums[i + offset] == val:
            offset += 1
        else:
            nums[i] = nums[i + offset]
            i += 1

    return len(nums) - offset

Test

>>> from RemoveElement import removeElement
>>> nums = [3, 2, 2, 3]
>>> k = removeElement(nums, 3)
>>> nums[:k]
[2, 2]
>>> nums = [0, 1, 2, 2, 3, 0, 4, 2]
>>> k = removeElement(nums, 2)
>>> nums[:k]
[0, 1, 3, 0, 4]

RemoveElement.removeElement(nums: List[int], val: int) → int: Removes all instances of val from the array nums in-place.