Binary Array Sorting
A binary array is an array consisting of only 0s and 1s.

You are given a binary array "arr" of size ‘N’. Your task is to sort the given array and return this array after sorting.

Input Format :

The first line of input contains an integer ‘T’ denoting the number of test cases. The next ‘2*T’ lines represent the ‘T’ test cases. The first line of each test case contains an integer ‘N' denoting the size of the array. The second line of each test case contains ‘N’ space-separated integers consisting of 0s and 1s as the array elements.

Output Format :

For each test case, print ‘N’ space-separated integers representing the elements of the sorted binary array in a separate line. Print the output of each test case in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Follow Up :

1. Can you solve this problem in a linear time and constant space? 2. Can you solve this problem in a single traversal only?

Constraints :

1

Question

Binary Array Sorting

A binary array is an array consisting of only 0s and 1s.

You are given a binary array "arr" of size ‘N’. Your task is to sort the given array and return this array after sorting.

Input Format :

 The first line of input contains an integer ‘T’ denoting the number of test cases.

 The next ‘2*T’ lines represent the ‘T’ test cases.

 The first line of each test case contains an integer ‘N' denoting the size of the array.

 The second line of each test case contains ‘N’ space-separated integers consisting of 0s and 1s as the array elements.

Output Format :

For each test case, print ‘N’ space-separated integers representing the elements of the sorted binary array in a separate line.

Print the output of each test case in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Follow Up :

1. Can you solve this problem in a linear time and constant space? 
2. Can you solve this problem in a single traversal only?

Constraints :

1 <= T <= 50
1 <= N <= 10^4
0 <= arr[i] <= 1

Where ‘T’ is the number of test cases and ‘N’ is the size of the array.

Time Limit: 1 sec

Accepted Answer

A two-pointer approach can be used for this question. Maintain two indexes. Initialize the first index left as 0 and second index right as n-1, where n is size of the array .
While left < right , do the following :
a) Keep incrementing index left while arr[left] =0
b) Keep decrementing index right while arr[right]=1
c) If left < right then exchange arr[left] and arr[right]

Accepted Answer

Counting

Initialize an integer variable ‘countZero’:= 0, It will store the count of zeros in the given array.
Iterate over the given array, if the current element is 0 then increment ‘countZero’ by 1.
Run a loop where ‘i’ ranges from ‘0’ to ‘n-1’. If ‘i’ < ‘countZero’ then place 0 at that index in the given array, otherwise place 1.
Return this array.

Note -: Instead of counting 0 we can also count 1 in the given array and similarly solve this problem.

Space Complexity: O(1)Explanation:

O(1)

We are not using any extra space here.

Time Complexity: O(n)Explanation:

O(n), where ‘n’ is the size of the array.

We are visiting every element of the array twice at max, hence making the complexity O(n).

Accepted Answer

One pass of quick Sort

This problem can be solved in a single traversal also. The concept used for this approach is related to the partition of quicksort. In a quick sort’ partition, after one scan, the left of the array is the smallest and the right of the array is the largest of the selected pivot element. Here we don't make any pivot but we make the smaller(0) come back in the array and Larger value(1) go ahead in the array.

Create an integer variable ‘index’ = 0.
Traverse the array from start to end and do the following -
- If the element is 0, then swap the current element with the element at ‘index’ position. and increment the ‘index’ by 1.
- If the element is 1 keep the element as it is.
Return this array.

Space Complexity: O(1)Explanation:

O(1)

We are using constant space.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the size of the array.

We are doing single traversal of the given array, hence making the complexity O(N).