This set contains practice questions on augmented data structures and on priority queues.
Question 1  Searchable stack. The ADT searchable stack is supposed to handle the following operations on on sets of data with two keys (for example, student name, and student ID number). The operations are PUSH, POP, TOP, MAKENULL, SEARCHBYKEY1, and SEARCHBYKEY2. Using the principle of augmentation of a data structure, suggest a data structure for this situation, such that PUSH, POP and both SEARCH operations take time O(log n) (n is the number of elements in the data structure at the time of the call), and the other operations take time O(1). 
Answer 
Because we must search, it seems natural to augment
a redblack search tree. One possible answer is to use
a redblack tree for each of the two keys, with
nodes containing the following information fields:

Question 2  Maxstack. The ADT maxstack is like an ordinary stack, with an extra operation, MAX, which returns the maximum of the key values of the elements in the stack. Show an implementation that allows all operations to be performed in O(1) worstcase time. 
Answer 
In an ordinary linkedlist based stack, we keep
a pointer "nextbelow" as in a single linked list.
However, if we keep a second pointer with each node,
called "maxbelow", which points to the node with maximal
key value at or below the given node, then all
operations can be done in time O(1).
Thus, a node contains the fields
nextbelow
if key[newnode]>key[maxbelow[TOP]] then maxbelow[newnode]=newnode else maxbelow[newnode]=maxbelow[TOP] TOP=newnode 
Question 3 
The ADT medianpriorityqueue (or MPQ for short).
The ADT medianpriorityqueue operates on a set of
elements with different key values, and
involves the following operations: makenull, insert, deletemedian,
makeMPQ.
It is known that the number of elements never exceeds a
given number n.
The operation makeMPQ takes a given number of
elements and inserts these globally (not one by one; recall
buildheap for a similar situation).
For deletemedian, note that
the median of a set of 2n elements is the nth smallest,
and that the median of a set of 2n+1 elements is the n+1st
smallest element.
Suggest, adapt or invent a data structure so that the
following can be achieved:

Answer  Among the possible solutions, the most elegant one uses two binary heaps of equal size, HL and HR (L for left, R for right). HL is a maxheap, HR is a minheap, and all elements of HL are smaller than those of HR. Furthermore, if . denotes size, HL = HR, or HL=HR+1. To insert newnode, insert it in HL if key[newnode] < key[min[HR]] (in O(log n) time of course). Insert it in HR otherwise. Note that the median is always the node max[HL] (the root of HL), so it can be deleted in O(log n) time. After "insert" and "delete", the balance condition may be violated. But by transferring max[HL] to HR or min[HR] to HL, we may reestablish it. This requires at most one additional deletemin/deletemax and one additional heapinsert. The total time is O(log n). The operation MakeMPQ involves finding the median in O(n) time by a medianselection algorithm, and then running "buildheap" on the sets of elements at or below the median (to make HL) and on the elements above the median (to make HR). 
Question 4  kary heaps. A kary heap is a generalization of a standard binary heap so that there are k children per node. Give an implicit data structure implementation, with details on how to find the ith child of node j, and the parent of node j. Estimate the worstcase number of comparisons (as a function of n and k, where n is the number of elements) for a standard operation DELETEMIN. Same question for the operation INSERT. 
Answer  Clearly, one should use an array in which nodes are kept in level order. If we begin with a root of index 0, then the children of a node with index i have indices ki+1 through ki+k. And the parent of a node with index i has index floor((i1)/k). The complete kary tree has height about log_k n. For a deletemin, we need k comparisons to descend one level, so the number of comparisons in the worst case is about k log_k n. For an insert, only one comparison per level is required, for a worstcase total of about log_k n. (Note that inserts are thus "cheaper" than deletemins). 
Question 5  kary heaps (continued). Let a kary heap with n elements be given. Let the number of DELETEMIN and INSERT operations be equal; the sequence of the operations is such that the size of the heap remains about n. Argue why a 4ary heap is best (you may need a calculator). 
Answer  By what we said in the previous answer, it is best to minimize (k+1) log_k n = ((k+1)/log k) log n. Now, take a calculator and compute (k+1)/log k, and notice that the values at k=2, 3, 4, and 5 are 4.32, 3.64, 3.60, and 3.72 respectively, and that these values increase from k=5 onwards. Thus, k=4 is the best choice! 
Question 6 
Beap.
In 1976, Munro and Suwanda proposed the beap
(biparental heap) as a possible implementation for
a priority queuelike ADT supporting the operations
makenull, insert, deletemin and search.
A beap of
n elements should be visualized as a triangular matrix such
as the one shown below, which has 10 elements.
The elements are added in crossdiagonal fashion,
so that the inherent order is 1, 2, 4, 6, 5, 9, 21, 8, 20
and 13 in the example shown below.
1 4 9 13 ... 2 5 20 ... 6 8 ... 21 ... 
Answer  Of course, one could use a double array of kxk elements, where k = 1 + sqrt(2n). However, as half the array is necessarily unused, one could also use a linear array, where elements are placed in the order (0,0), (0,1),(1,0), (0,2),(1,1),(2,0), and so forth. In both cases, it is necessary, given n, to determine the nth element in the structure. In the former case, neighbors in the structure are easy to find, while in the latter, some work is needed. Let us number the diagonal rows of elements, and keep track of the highest index on each diagonal. For the ith diagonal, the last element has index 1 + 2 + ... + i = i(i+1)/2. Let i = diagonal(n) be the index of the diagonal containing n. This index is O(sqrt(n)) and diagonal(n) takes at most O(sqrt(n)) time to compute. indeed: given n, we find the unique i for which (i1)i/2 < n <= i(i+1)/2 by verifying the inequalities starting with i=1. The ith diagonal has i elements elements numbered (0,i1), (1,i2), ... , (i1,0). The nth element overall is thus at array position (i1m,m) where m = i(i+1)/2 n. 
Question 7  Beap (continued). Inspired by similar operations on heaps, describe how you would implement makenull, insert, search, and deletemin. 
Answer 
Let us use the array approach.
Assume that the beap holds n elements and that
x is the key value of a new element to be inserted.
We first fix the shape,
by placing the new element at position n+1 (see
previous question on how to do this in O(sqrt(n))
time).
Repeat forever:
If (c',r') = (c,r) then stop. Else exchange the keys of (c,r) and (c',r'), and set (c,r) < (c',r').
Set (c,r) = (0,0) Repeat forever:
If (c',r') = (c,r) then stop. Else exchange the keys of (c,r) and (c',r'), and set (c,r) < (c',r'). 
Question 8 
Beap (continued).

Answer  The first subquestion was addressed in my previous answer: it isd O(sqrt(n)). Beapsort takes time O(n^(3/2)) as each of the n inserts and n deletemins takes time bounded by a constant times n^(1/2). For a future event set, inserts and deletemins are equally frequent. Clearly, binary heaps are to be preferred. Finally, of the four data structures listed, the first group of operations is best handled by a binary heap (in time O(n log n)), the second one by the sorted array (in time O(log n)), and the third one by a beap (in time O(n^(3/2))) (the three other structures take worstcase time Theta (n^2) in the latter case). 
Copyright © 1999 Luc Devroye. Email: luc@cs.mcgill.ca.