Remove Element | Leetcode #27 | Easy

Rishikesh Dhokare
3 min readJan 9, 2021


In this post I will discuss the solution to the leetcode problem — Remove Element


Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.


Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {

Example 1:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length. For example if you return 2 with nums = [2,2,3,3] or nums = [2,2,0,0], your answer will be accepted.

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 length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4. Note that the order of those five elements can be arbitrary. It doesn't matter what values are set beyond the returned length.


  • 0 <= nums.length <= 100
  • 0 <= nums[i] <= 50
  • 0 <= val <= 100


This is very similar to one of the previous problems on leetcode i.e. Remove Duplicates from Sorted Array. The only difference is, instead of removing all the duplicates, we need to remove a specified value i.e. val

So the algorithm remains almost the same except the comparison. These are the steps —

  1. As a base condition, if the nums array is empty, we can straightaway return 0 as output.
  2. We can iterate the array with 2 pointers. Let’s call them slow and fast and initialize both of them to 1 as we will start the array iteration from the second element of the array.
  3. While iterating the input array, we compare current element with val.
  4. If current element is not equal to val, we store the element at index fast to the index of slow and increment slow by one. Otherwise slow remains the same.
  5. Return slow as the output as that is the count of distinct elements in the array.

This is how the code looks like

class Solution {
public int removeElement(int[] nums, int val) {
if (nums.length == 0) {
return 0;
int slow = 0;
for (int fast = 0; fast < nums.length; fast++) {
if (nums[fast] != val) {
nums[slow] = nums[fast];
return slow;

Hope this helps! Happy coding! 🙂

If you think the solution can be improved or misses something, feel free to comment. There is always some room for improvement.

Find the solutions to the leetcode problems here —



Rishikesh Dhokare

I am a Software Engineer from India and working in Berlin, Germany. I write about technology, my experiences in Germany, travel in Europe.