Question

This is not a homework. Just an interesting task 🙂

Given a complete binary search three represensted by array. Sort the array in O(n) using constant memory.

Example:

Tree:

              8
           /     \
          4       12
         /\       / \
        2  6     10  14
       /\  /\    /\   /\
      1 3 5  7  9 11 13 15

Array: 8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15

Output: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Answer 1

It is possible, people calling it homework probably haven’t tried solving it yet.

We use the following as a sub-routine:

Given an array a1 a2 ... an b1 b2 .. bn, convert in O(n) time and O(1) space to

b1 a1 b2 a2 ... bn an

A solution for that can be found here: http://arxiv.org/abs/0805.1598

We use that as follows.

Do the above interleaving for the first 2^(k+1) – 2 elements, starting at k=1 repeating for k=2, 3 etc, till you go past the end of array.

For example in your array we get (interleaving sets identified by brackets)

 8, 4, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15   
[ ][ ]

 4, 8, 12, 2, 6, 10, 14, 1, 3, 5, 7, 9, 11, 13, 15   (k = 1, interleave 2)
[        ][        ]  

 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15   (k = 2, interleave 6)
[                      ][                     ]

 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15   (k = 3, interleave 14)

So the total time is n + n/2 + n/4 + … = O(n).
Space used is O(1).

That this works can be proved by induction.