LeetCode 27 Remove Element

Description

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

assert k == expectedLength;
for (int i = 0; i < k; 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, 1, 4, 0, and 3.
Note that the order of those five elements can be arbitrary.
It does not matter what values are set beyond the returned k (hence they are underscores).

Constraints:

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

We need to filter out all values equal to val, shifting valid ones to the front. Since the order doesn't matter, we can use a two-pointer approach:

The Right pointer move from right to left, skip val, point to last element that not val. The Left pointer move from left to right, if it point to val, replace with value of right pointer.

var removeElement = function(nums, val) {
    let left = 0
    let right = nums.length - 1
    while(left <= right) {
        if(nums[right] === val) {
            right--
        } else if (nums[left] === val) {
            nums[left] = nums[right]
            right--
            left++
        } else {
            left++
        }
    }
    return left
};