Mock interview: shuffle array with negative number together keep order, same as positive numbers

Here is the gist.

	5:34 PM
	Input: [1,2,-1,8,1,-1]
	Output: [-1,-1,1,2,8,1]
	Is [-1, -1, 1,1 ,2 ,8] acceptable?
	Brute force - go through twice, first time you go through negative number, put into the list, second time you go through the positive number, append -> list
	O(N), space O(N) list -> array - new list
	In place - O(N) time complexity, O(1) space
	Keep the order
	1, 2, -1, 8, 1, -1
	-1, 2, 1, 8, 1, -1
	-1, -1, 1, 8, 1, 2 -> 8, 1, is not position properly - swap
	From consider left to right
	1, 2, -1, 8, 1, -1
	I
	1, 2, -1, 8, 1, -1
	J
	1, 2, -1, 8, 1, -1
	i
	1, 2, -1, 8, 1, -1
	J
	1, 2, -1, 8, -1, 1
	i j
	1, 2, -1, -1, 8, 1
	i j
	1, -1, -1, 2, 8, 1
	i j
	-1, -1, 1, 2, 8, 1
	I = 1 is leftmost one to terminate the search
	-1 <= elements <= 10^5
	// 1, 2, -1, 8, 1, -1 < pos = 5, i = 5
	// 1, 2, -1, 8, 1, -1 < pos = 5, i = 4 - pos--
	// 1,2,-1, -1,8, 1 < pos = 4, i = 3
	// 1, -1, -1, 2,8, 1 < pos = 3, i =2 i: pos:
	// 1, -1, -1, 2, 8, 1 < pos = 2, i = 1
	// -1, -1, 1, 2, 8, 1 < pos = 1, i = 0
	283. Move Zeroes
	Void MoveNegativeOneToFrontKeepOrder(int[] numbers) // -1, 3, 2
	{
	if(numbers == null ||numbers.Length == 0)
	return;
	// count -1 number in total
	// put all positive from last one to the end - do one pass
	// var countNegative = 0;
	var length = number.Length;
	var pos = length - 1; // 3
	for(int i = length - 1; i >= 0; i--)
	{
	var current = numbers[i]; // 2
	Var isNegative = current == -1;
	if(!isNegative)
	{
	numbers[pos] = current; // numbers[2] = 2
	// numbers[i] = -1; // replace its value
	pos--;// 1
	}
	}
	}
	public class Solution {
	public void MoveZeroes(int[] nums)
	{
	if (nums == null)
	return;
	var nonZeroCount = 0;
	var length = nums.Length;
	for(int i = 0; i < length; i++)
	{
	var current = nums[i];
	if (current == 0)
	continue;
	nums[nonZeroCount] = current;
	nonZeroCount++;
	}
	for(int i = nonZeroCount; i < length; i++)
	{
	nums[i] = 0;
	}
	}
	}

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