# Day 2 - Remove Element - Leetcode Problem Solved

## #I4G10DaysOfCodeChallenge - Day 2

In this article, I share my solution and thought process for day 2's Leetcode challenge for the #I4G10DaysOfCodeChallenge series. I have actually also solved this problem on Leetcode before, but having another go at it made me realize a faster solution.

Let's goo...

*Problem statement, examples, and constraints.*

*Problem statement, examples, and constraints.*

## Remove Element

### #I4G10DaysOfCodeChallenge

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 = [0,0,1,1,1,2,2,3,3,4] Output: 5, nums = [0,1,2,3,4,_,_,_,_,_] Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).

**Constraints:**

`0 <= nums.length <= 100`

`0 <= nums[i] <= 50`

`0 <= val <= 100`

**Solution**

```
var removeElement = function(nums, val) {
for(let index = 0; index < nums.length; index++) {
if(nums[index] === val) {
nums.splice(index, 1);
index--;
}
}
return nums.length;
};
```

*Thought Process*

*Thought Process*

From the problem statement, the whole idea of this problem is to remove all occurrences of `val`

**in-place**. This is a relatively simple problem because I don't have to keep the relative order of the elements the same.

Since this operation has to be done in place, it's somewhat impossible to change the length of the array. Hence, the reason why my result has to occupy the first parts of the array it requires.

I use a very straightforward approach in solving this problem, All I'm doing is this; I'm looping through the array, and checking each element to see if it matches with the value I'm searching for. If an element matches, I'll remove the element and continue the search to make sure all occurrences of that value are removed. But if none of them form a match, I'll just return the length of the original array.

```
for(let index = 0; index < nums.length; index++) {
if(nums[index] === val) {
nums.splice(index, 1);
index--;
}
}
return nums.length;
```

*Efficiency*

*Efficiency*

With regard to efficiency, I think this is quite an optimal solution. Time Complexity: **O(n)** Space Complexity: **O(1)**

**Closing**

Thanks for reading through, I hope you learned a thing or two. Will see you in the next challenge post.

Please share your thoughts in the comment section, let's talk :-)

Your friend in progress,

*CodeProphet.*