LeetCode 27

27. Remove Element

팁

Clicking the heading will take you to the LeetCode problem.

Solution

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        # Initialize a pointer to track the index for non-val elements 
        i = 0

        # Iterate through the array
        for j in range(len(nums)):
            # If the current element is not equal to val
            if nums[j] != val:
                # Place the element at the i-th position
                nums[i] = nums[j]
                # Move the pointer i forward
                i += 1
        
        # Return the length of the modified array
        return i

The time complexity is $O(n)$ because it iterates through the length of the array. Since the operation is performed in-place, no additional space is required. Therefore, the space complexity is $O(1)$.