27. 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 arrayin-place with O(1) extra memory.

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

Clarification:

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++) {
    print(nums[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 = [2,3,0,0], 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.

Constraints:

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

# @lc code=start
using LeetCode

remove_element!(nums::Vector{Int}, val::Int) = length(filter!(!=(val), nums))
# @lc code=end

remove_element! (generic function with 1 method)

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