How do you solve Remove Element on LeetCode?

Remove Element (LeetCode #27) can be solved using Brute Force or Optimal Solution (Two Pointers). The optimal approach is Optimal Solution (Two Pointers) with O(n) time complexity and O(1) space complexity. Key insight: Using the two pointers technique allows us to efficiently modify the array in-place without needing extra space.

What is the brute force solution for Remove Element?

The brute force approach for Remove Element is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a new array and copying over all elements that are not equal to the specified value. This is straightforward but inefficient as it requires extra space and multiple passes through the array. The optimal Optimal Solution (Two Pointers) approach improves this to O(n).

What algorithm or data structure is used to solve Remove Element?

Remove Element uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Remove Element?

When you first see this problem, clarify that you understand it requires in-place modification and that the order of elements can change. Communicate your approach clearly, mentioning that you plan to use the two pointers technique to track valid elements. Be sure to mention edge cases, such as when the array is empty or when all elements are equal to val.

What are common mistakes when solving Remove Element?

Students often try to create a new array instead of modifying the original array in-place, leading to unnecessary space usage. Another common mistake is not properly managing the index of the valid elements, which can lead to incorrect counts.

#27

Remove Element

Easy

ArrayTwo PointersHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Two Pointers)★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses the two pointers technique to efficiently remove elements in-place. One pointer traverses the array while the other keeps track of the position to place the next valid element.

⚙️

Algorithm

4 steps

1Step 1: Initialize a pointer 'k' to 0, which will track the position of the next valid element.
2Step 2: Iterate through the array with another pointer 'i'.
3Step 3: If nums[i] is not equal to val, assign nums[k] = nums[i] and increment k.
4Step 4: After the loop, k will be the count of elements not equal to val.

solution.py9 lines

1# Full working Python code with comments
2
3def removeElement(nums, val):
4    k = 0  # Pointer for the next valid position
5    for i in range(len(nums)):
6        if nums[i] != val:
7            nums[k] = nums[i]  # Place valid element at position k
8            k += 1  # Increment k
9    return k  # Return the count of valid elements

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the array, and the space complexity is O(1) since we are modifying the array in place without using additional storage.

1Using the two pointers technique allows us to efficiently modify the array in-place without needing extra space.
2Understanding that we can overwrite elements in the original array rather than removing them can simplify the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.