## Accolite software Interview Question

**Country:**United States

**Interview Type:**In-Person

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--;
}
}
}
```

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;
}
```

@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.

- Alexey July 23, 2016