27. Remove Element

Question

Given an integer array nums and an integer val , remove all occurrences of val in nums in-place . The order of the elements may be changed. Then return the number of elements in nums which are not equal to val .
Consider the number of elements in nums which are not equal to val be k , to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the elements which are not equal to val . The remaining elements of nums are not important as well as the size of nums .
Return k .
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).

Constraints:

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

Solutions

// two pointer approach:
// - slow points to the position where need to be replaced or not.
// - fast iterates through the array. If nums[fast] is not equal to val, then replace it with nums[slow]
// [0, left) ensures no val.
// edge cases:
// 1. when nums is empty -> no need to remove, would skip and return 0
// 2. when nums only have one element -> follow the for loop 
// 3. when all elements in nums are equal to val -> slow keeps 0, and no override op.
// 4. when no elements in nums are equal to val -> do N times overrides and slow increment to the size of array.
// Time Complexity: O(N)
// Space Complexity: O(1)
int removeElement(vector<int>& nums, int val) {
	int n = static_cast<int>(nums.size());
	int slow = 0;
	for (int fast = 0; fast < n; ++j) {
		if (nums[fast] != val) {
			nums[slow] = nums[fast];
			++slow;
		}
	}
	return slow;
}