LeetCode: Remove Element

Problem:

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

Code:

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        j=0 # write pointer
        for i in range(len(nums)): # i is read pointer
            if nums[i]!=val:
                nums[j]=nums[i]
                j+=1
        return j

Key Points:

• Goal: remove all occurrences of val in-place and return the new length.

• Use two pointers: i → read pointer (checks every element) j → write pointer (builds the cleaned list)

• For each element:

  • If nums[i] != val:

    • Copy nums[i] to nums[j]

    • Move j forward

  • If nums[i] == val:

    • Skip it (do not copy)

• After the loop, j = number of kept elements.

• The array's first j positions now contain the valid filtered values.

• Time: O(n) because we scan once.

• Space: O(1) because we modify in-place without extra storage.