Remove Element LeetCode solution


Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

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

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                            // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
    assert nums[i] == expectedNums[i];

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).

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 k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).


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


The problem looks pretty simple to understand. But what makes it tricky is that we have to do this in place. My initial thought process was that I use a bitmap and then hold the index of all the positions of to be removed element. And then use the bitmap to iterate these positions to the end of the array. But this would be very inefficient (Big O N*N complexity).

There is an easier intuitive solution, although I would admit, it didn’t strike easily. It goes like this. You maintain an index variable (initially initialized to 0, beginning of the array). And you iterate through the array, and whenever you find an element which is not equal to value to be removed, then you write the value at the index and increment index by one. This way once you are done with the array, all the non value elements would have been written continuously in the array. And the index represents length of this continuous array.

Below is the Java code for the same.

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

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