A heaping challenge. We begin by calling function solution(Heap h, int x, int k) with the heap we’d like to search h, the data item x we’re comparing with, and the number of items k that we’d like to see less than x. solution then calls solveRecursive(HeapNode i, int x, int k), passing the root node of h, as well as x and k. solveRecursive then follows the following logic:

1. If i is null, or k is 0, or the value at i is not less than x, return 0.
2. Else, if i has no children, return 1.
3. Else, call solveRecursive, passing parameters i.left (the left child of i), x, and k - 1, storing the result as lsum.
4. If lsum is greater than or equal to k - 1, then return lsum + 1.
5. Else, call solveRecursive, passing parameters i.right (the right child of i), x, and k - 1 - lsum, and store the result as rsum.
6. Return 1 + lsum + rsum

Then, in solution, return true if the final result is greater than or equal to k, and false otherwise.

As pseudocode, this could be written as:

 #define true 1
 #define false 0
 typedef int bool;
 typedef unsigned int uint32_t;

 bool solution(Heap h, int x, uint32_t k) {
   return solveRecursive(h.root, x, k) >= k;
 }

 uint32_t solveRecursive(HeapNode i, int x, uint32_t k) {
   if (i == NULL || k <= 0 || i.value >= x)
     return 0;
   else if (i.left == NULL && i.right == NULL) {
     return 1;
   } else {
     uint32_t sum = solveRecursive(i.left, x, k-1) + 1;
     if (sum >= k) return sum;
     else return sum + solveRecursive(i.right, x, k - sum);
   }
 }

We first define a list-backed heap node to be a heap node with no children and a reference to a sorted list, and whose delete operations replace the node’s current value with the next value from the list (by index), unless the list is empty, in which case the node is successfully deleted. In both cases the node’s previous value is returned.

We can then create a list-backed heap node from each of the k lists, and combine them pairwise into a single heap using this process:

1. Given heaps h1 and h2, delete min from the heap with the smaller root node value, storing the result as r.
2. Create a heap with root node with value r, and children h1 and h2.

Note that this process doesn’t maintain the shape of the resulting heap; the final heap is not guarranteed to be a complete tree. This operation is done k - 1 times to combine all of the list-backed heap nodes into a single heap of 2k - 1 nodes, and each operation has a runtime of on average O(log k), meaning that this process is O(k*log k). We then output the result of n delete min operations from the final heap we have created. each delete min is a O(log k) operation, meaning that this operation is O(n*log k), and the whole process is O((n + k)*log k). Since n ≫ k, this simplifies to O(n*log k).

We use a modified version of a 2-3 tree, where each guide now describes the total number of elements contained in its subtree, and each value is an element in a list. The values are stored in the order that they appear in the list. Then the following operations can be performed

• O(1) List creation: To create a list, we instantiate a new 2-3 tree of known size. This is a constant time operation.
• O(log n) List Concatenation: To concatenate two lists, L₁ and L₂, we follow the general procedure for combining 2-3 trees, with 3 cases:
  1. L₁ has height greater than L₂: We find some node p on the right-most path of L₁ whose subtree has height one greater than the height of L₂. We then perform an insert, inserting L₂ as the right-most child of p. This insert is done identically to a general ordered 2-3 tree, except that each time we try to insert a child, we also update the parent’s guide to be the sum of all its childrens’ guides.
  2. L₂ has height greater than L₁: This reduces to the previous case, except that we define p as a node on the left-most path of L₂ whose subtree has height one greater than the height of L₁, and we insert L₁ as the left-most child of p.
  3. L₁ has the same height as L₂: We return a new list whose children, in order, are L₁ and L₂, and whose guide is the sum of the guides of the root nodes of those two lists.
• O(log n) List Splitting: To split a list, follow the procedure that you would for an ordered 2-3 tree, substituting the above list concatenation method for the ordered 2-3 tree method. To traverse to the k^th item, use the next item.
• O(log n) Get the k^th Item: To get the k^th item, use the following procedure:
  1. Begin with some integer i = k, and node n = RootNodeOfL, with guide n.guide and children n.child0, n.child1, and n.child2, where n.child2 is undefined if n has only 2 children.
  2. If i > n.guide, then k is not a valid index. Return an error code.
  3. If n has the guide 1, n is a leaf node! Return the value at n.
  4. If n.child0 has a guide greater than or equal to i, set n to n.child0, and go back to 3.
  5. If n.child0.guide + n.child1.guide is greater than or equal to i, set n to n.child1, and go back to 3.
  6. Set n to n.child2 and go back to 3.

To accomodate for fast reversing of L, we add a reverse bit to the guide of each node in L, which by default is set to 0. We then define two modes of reading/writing to the structure of L:

• Normal Mode: The children are in order from left to right (as before). When a node has 2 children, its child2 is undefined
• Reversed Mode: The children are in reverse order. The right-most child is child0, and the left most child is child1 when the node has 2 children and child2 when the node has 3 children.

When performing any of the operations described in problem 5, we start with our read/write mode as normal mode, and switch between normal and reversed mode whenever we come accross a node with a reversed bit of 1. For example, if L has exactly one reverse bit set to 1 in its root, then we interpret every node in reverse mode. When finding the k^th element of L, we traverse using the same method as above, but do the operations from right to left instead of left to right if we’re in reverse mode, i.e. do step 4 with n.child2 (if its defined), step 5 with n.child2 and n.child1 (if n.child2 is defined), and so on.