CareerCup

function sort01(array) {
    if (!array || array.length < 2) {
        return;
    }

    var firstNonZeroIndex = null;
    var index = 0;

    while(index < array.length) {
        if (array[index] === 1 && firstNonZeroIndex === null) {
            firstNonZeroIndex = index;
        }

        if (array[index] === 0 && firstNonZeroIndex !== null) {
            [array[index], array[firstNonZeroIndex]] = [array[firstNonZeroIndex], array[index]];
            firstNonZeroIndex++;
        }

        index++;
    }
}

var a = [1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0];
sort01(a);
console.log(a);

Use partition function of Quicksort, easy-peasy.

public static void main(String[] args) {
		int arr[] = { 0, 1, 1, 1, 1, 0, 0, 1 };
		for (int i = 0, j = arr.length - 1; i < j; i++, j--) {
			if (arr[i] > arr[j]) {
				int tmp = arr[i];
				arr[i] = arr[j];
				arr[j] = tmp;
			}
		}
                //print the arr 
		for (int i : arr) {
			System.out.println(i);
		}
	}

The requirements for
1. O(n)
2. In place sort
3. preserve order (stable)
are a stingy combination.

All common sorting algorithms takes O(nlogn):
-Quicksort can sort in-place, but its partitioning function is not stable (does preserve element order), and it's not O(n)
-Mergesort can sort in-place, and preserve order (stable), but is not O(n)
-Bubble sort sort in-place, but O(n^2)

Radix sort is possibly the only sub-O(nlogn) algorithm. It runs O(wn) where w is the length of word (in this problem where elements are either 1 or 0, w=1). It also preserves order, but does not sort in place (it creates bucket).

None of the solution offered so far satisfy the 3 requirements simultaneously. I think interviewer is looking for RadixSort, which gives O(n) and stable order preservation. But I doubt he meant in-place (otherwise, creating buckets during RadixSort cost extra space)

O(n) runtime. the basic concept is to keep swapping at the beginning of the array where the 0's are untill that is filled.

def stable_sort(array):
	j_start = 0
	for a in array:
		if a == 0:
			j_start += 1
			
	i_start = 0
	j_end = j_start
	
	while i_start != j_start:
		for i in range(i_start, j_start):
			if array[i] != 0:
				array[j_end], array[i] = array[i], array[j_end]
				j_end += 1
			else:
				i_start += 1
	
	return array

The basic concept is to move out the 1's from the area that the 0's will occupy. By always swapping into the bucket where the 0's will occupy, you can preserve the order.

here's an example of a run through the algorithm:

1a 1b 1c 1d 0a 0b 1e 1f
j_start = 2
j_end = 2
i_start = 0

i=0, j_end=2
1c 1b 1a 1d 0a 0b 1e 1f

i=1, j_end=3
1c 1d 1a 1b 0a 0b 1e 1f

i=0, j_end=4, i_start = 1
0a 1d 1a 1b 1c 0b 1e 1f

i=1, j_end=5, i_start = 2
0a 0b 1a 1b 1c 1d 1e 1f

i_start == j_start so stop.

def stable_sort(array):
	j_start = 0
	for a in array:
		if a == 0:
			j_start += 1
			
	i_start = 0
	j_end = j_start
	
	while i_start != j_start:
		for i in range(i_start, j_start):
			if array[i] != 0:
				array[j_end], array[i] = array[i], array[j_end]
				j_end += 1
			else:
				i_start += 1
	
	return array

What does it mean "While sorting you are not allowed to change the original ordering of same element" - do you mean sort must be stable?
So you want stable in-place partition? Sorry, but that doesn't sound like an interview question.
citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.25.5554

void swap(vector<int>& arr, int one_pos, int zero_pos)
{
    int tmp = arr[one_pos];
    arr[one_pos] = arr[zero_pos];
    arr[zero] = tmp;
}

void sortInPlace(vector<int>& arr)
{
    int zero_pos = arr.size()-1;
    for (int i = arr.size()-1; i>=0; i--)
    {
        if (arr[i] == 0 && arr[zero_pos]!=0)
            zero_pos = i;
        else if (arr[i] == 1 && arr[zero_pos] == 0)
        {
            swap(arr, i, zero_pos);
            zero_pos--;
        }
    }
}

If the sort starts from the end, it can keep the 0s and 1s in their original order. That means the last 1 should be at the end and the first 0 should be at the beginning. Here is the function.

void swap(vector<int>& arr, int one_pos, int zero_pos)
{
    int tmp = arr[one_pos];
    arr[one_pos] = arr[zero_pos];
    arr[zero_pos] = tmp;
}

void sortInPlace(vector<int>& arr)
{
    int zero_pos = arr.size()-1;
    for (int i = arr.size()-1; i>=0; i--)
    {
        if (arr[i] == 0 && arr[zero_pos]!=0)
            zero_pos = i;
        else if (arr[i] == 1 && arr[zero_pos] == 0)
        {
            swap(arr, i, zero_pos);
            zero_pos--;
        }
    }
}

print stable_sort([1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0])

[0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0]

next time, test your code before you post

Instead of using two pointers, start and end which will modify the order, use start and middle so the order is preserved

int[] preservedSort(int[] array)
{
	if (array == null || array.length() == 0)
		return array;
	int start = array[0];
	int middle = array.length() / 2 ;
	while(start < array.length() && middle < array.length())
	{
		if (array[start] > array[middle])
		{
			swap(array[start], array[middle]);
			start++; middle++;
		}
		if (array[start] == 0)	start++;
		if (array[middle] == 1) middle++;
	}

	return array; 
}

public static int[] SortOsAnd1sInPlaceO_n(int[] n)
        {
            int i=0, j=n.Length-1;
            while(i < j)
            {
                if(n[i] > n[j])
                {
                    int temp = n[j];
                    n[j] = n[i];
                    n[i] = temp;
                    i++;
                    j--;
                }
                else if(n[i] < n[j])
                {
                    i++;
                    j--;
                }
                else
                {
                    i++;
                }
            }
                return n;
        }

@emb and all, this question becomes a lot easier if you consider that using extra space is permitted. Why anybody would want an in-place sorting while allowing extra space is beyong me, though... but a solution that fulfills all the requirements is much easier to find that way. I say this because I think many people here implicitly assume no extra space is permitted.

while(start < end)
{
while(arr[start] ==0)
start++;
while(arr[end]== 1)
end++;
swap(start,end);
}

int arr[] = { 0, 1, 1, 1, 1, 0, 0, 1 };
int x =arr.length;
for (int i = 0; i < arr.length/2; i++) {
if(arr[i]>arr[x-1]){
int temp = arr[x-1];
arr[x-1] = arr[i];
arr[i] = temp;
}
x--;
}
for (int i = 0; i < arr.length; i++) {
System.out.println(arr[i]);
}

int arr[] = { 0, 1, 1, 1, 1, 0, 0, 1 };
int x =arr.length;
for (int i = 0; i < arr.length/2; i++) {
if(arr[i]>arr[x-1]){
int temp = arr[x-1];
arr[x-1] = arr[i];
arr[i] = temp;
}
x--;
}
for (int i = 0; i < arr.length; i++) {
System.out.println(arr[i]);
}

Can't we just count the number of zeroes as 'x' and set the initial 'x' elements as 0 and the rest as 1. It will run in O(n) time.