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Problem 1. Binary Tree Practice
(a) The Set Binary Tree T below is not height-balanced but does satisfy the binary
search tree property, assuming the key of each integer item is itself. Indicate the
keys of all nodes that are not height-balanced and compute their skew.
Solution:
The nodes containing the keys 16 and 37 are not height balanced. Their skews are 2
and -2 respectively.
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(b) Perform the following insertions and deletions, one after another in sequence
on T, by adding or removing a leaf while maintaining the binary search tree
property (a key may need to be swapped down into a leaf). For this part, do not
use rotations to balance the tree. Draw the modified tree after each operation.
1 T.insert(2)
2 T.delete(49)
3 T.delete(35)
4 T.insert(85)
5 T.delete(84)
Solution:

insert(2)
delete(49)

delete(35)
insert(85)

delete(84) (Two solutions, swap down
predecessor/successor)
(c) For each unbalanced node identified in part (a), draw the two trees that result
from rotating the node in the original tree left and right (when possible). For each
tree drawn, specify whether it is height-balanced, i.e., all nodes satisfy the AVL
property.

Solution:
Node containing 16 is not height-balanced.
• Rotating left at this node does not balance the tree:
• Rotating right at this node does not balance the tree:

Node containing 37 is not height-balanced.
• Rotating left at this node is not possible
• Rotating right at this node results in a height-balanced tree!
Problem 2. Heap Practice
For each array below, draw it as a complete1 binary tree and state whether the
tree is a max-heap, a min-heap, or neither. If the tree is neither, turn the tree into
a min-heap by repeatedly swapping items that are adjacent in the tree.
Communicate your swaps by drawing a sequence of trees, marking on each tree
the pair that was swapped.

(a) [4, 12, 8, 21, 14, 9, 17]
Solution: Min-heap 4
(b) [701, 253, 24, 229, 17, 22]
Solution: Max-heap

(c) [2, 9, 13, 8, 0, 2]
Solution: Neither: three swaps suffice to transform into a min-heap
(d) [1, 3, 6, 5, 4, 9, 7]
Solution: Min-heap

Problem 3.
Gardening Contest Gardening company Wonder-Grow sponsors a nation-wide
gardening contest each year where they rate gardens around the country with a
positive integer2 score. A garden is designated by a garden pair (si, ri), where si is
the garden’s assigned score and ri is the garden’s unique positive integer
registration number.
(a) To support inclusion and reduce competition, Wonder-Grow wants to award
identical trophies to the top k gardens. Given an unsorted array A of garden
pairs and a positive integer k ≤ |A|, describe an O(|A| + k log |A|)-time
algorithm to return the registration numbers of k gardens in A with highest
scores, breaking ties arbitrarily.
Solution:
Build a max-heap from array A keyed on the garden scores si, which can be done
in O(|A|) time. Then repeatedly remove the maximum pair k times using delete
max(), and return the registration numbers of the pairs extracted (say in an array
of size k). Because a max-heap correctly removes some maximum from the heap
with each deletion in O(log |A|) time, this algorithm is correct and runs in O(|A| +
k log |A|) time.

(b) Wonder-Grow decides to be more objective and award a trophy to every
garden receiving a score strictly greater than a reference score x. Given a max-
heap A of garden pairs, describe an O(nx)-time algorithm to return the
registration numbers of all gardens with score larger than x, where nx is the
number of gardens returned.
Solution:
For this problem, we cannot afford O(nx log |A|) time to repeatedly delete the
maximum from A until the deleted pair has score less than or equal to x.
However, we can exploit the max-heap property to traverse only an O(nx) subset
of the maxheap containing the largest nx pairs. First observe that the max-heap
property implies all nx items having key larger than x form a connected subtree of
the heap containing the root (assuming any stored score is greater than x). So
recursively search node v of the heap starting at the root. There are two cases,
either:
• the score at v is ≤ x, so return an empty set since, by the max-heap property, no
pair in v’s subtree should be returned; or
• the score at v is > x, so recursively search the children of v (if they exist) and
return the score at v together with the scores returned by the recursive calls
(which is correct by induction).

This procedure visits at most 3nx nodes (the nodes containing the nx items
reported and possibly each such node’s two children), so this procedure runs in
O(nx) time.
Problem-4.
Solar Supply Entrepreneur Bonty Murns owns a set S of n solar farms in the town
of Fallmeadow. Each solar farm (si, ci) ∈ S is designated by a unique positive
integer address si and a farm capacity ci: a positive integer corresponding to the
maximum energy production rate the farm can support. Many buildings in
Fallmeadow want power. A building (bj , dj ) is designated by a unique name
string bj and a demand dj : a positive integer corresponding to the building’s
energy consumption rate.
To receive power, a building in Fallmeadow must be connected to a single solar
farm under the restriction that, for any solar farm si, the sum of demand from all
the buildings connected to si may not exceed the farm’s capacity ci. Describe a
database supporting the following operations, and for each operation, specify
whether your running time is worst-case, expected, and/or amortized.

Solution:
Our approach will be to maintain the following data structures:
• a Priority Queue P on the solar farms, storing for each solar farm its address si,
capacity ci, and its available capacity ai (initially ai = ci), keyed on available
capacity;
• a Set data structure B mapping each powered building’s name bj to the address
of the solar farm si that it is connected to and its demand dj ; and
initialize(S) Initialize database with a list S = ((s0, c0), . . . ,(sn1, cn1))
corresponding to n solar farms in O(n) time.
power on(bj,dj) Connect a building with name bj and demand dj to any solar
farm having available capacity at least dj in O(log n) time (or
return that no such solar farm exists).
power off(bj) Remove power from the building with name bj in O(log n)
time.
customers(si) Return the names of all buildings supplied by the farm at
address si in O(k) time, where k is the number of building
names returned.

• a Set data structure F mapping the address of each solar farm si to: (1) its own
Set data structure Bi containing the buildings associated with that farm, and (2) a
pointer to the location of si in P.
Now we support the operations:
• initialize(S): build Set data structures P and then F from S, and initialize all
other data structures above as empty. This operation directly maintains the
invariants of our database by reducing to build for F and P. There are O(n) empty
data structures and pointers constructed, so if we implement P and F with data
structures that can build in O(n) time, this operation will also take O(n) time.
• power on(bj, dj): assume that bj is not already connected to power (the
operation is unspecified otherwise). First, find a solor farm to connect by deleting
a solar farm si from P having largest available capacity ci (delete max) and
checking whether it’s capacity is at least dj . There are two cases:
– dj > ci, so reinsert the solar farm back into P (relinking a pointer from F to a
location in P) and return that no solar farm can currently support the building.
– dj ≤ ci, so subtract dj from ci and reinsert it back into P (relinking a pointer).
Then, add bj to B mapping to si, and then find the Bi in F associated with si and
add bj to Bi.

This operation directly maintains the invariants of our database and takes time
asymptotically upperbounded by the sum of: one delete max and one insert
operation on P, an insert operation on B, a find on F, an insertion into Bi, and
constant additional work (to maintain pointers and perform arithmetic).
• power off(bj): assume that bj is already connected to power (the operation is
unspecified otherwise). Lookup the si and dj associated with bj in B, lookup Bi in F
using si, and remove bj from Bi. Lastly, go to si’s location in P and remove si from
P, increase ci by dj , and reinsert si into P. This operation directly maintains the
invariants of our database, and takes time asymptotically upperbounded by the
sum of: one lookup in B, one lookup in F, one delete from Bi, one removal by
location from P, one insertion into P, and constant additional work.
• customers(si): lookup Bi in F using si, and return all names stored in Bi. This
operation is correct based on the invariants maintained by the data structure (Bi
contains the buildings associated with si), and takes time asymptotically
upperbounded by the sum of: one lookup in F and one iteration through Bi.
We have shown this database can correctly support the operations. Now we
choose implementations of the data structures in the database that will allow the
operations to be efficient. We need to be able to build B and F in O(n) time

with O(log n) lookups, so we must use hash tables for these, leading to expected
bounds on all operations, and amortized bounds on power on and power off. For
each Bi, we need O(log n) lookup, insert, and delete, so can implement with either
a Set AVL Tree or a hash table. Lastly, P requires O(n) build, and O(log n) insert
and delete max, so either a Max-Heap or a Sequence AVL Tree augmented with
subtree max items can be used. Note that removing an item from a Sequence AVL
Tree via a pointer to its node is exactly the same as deleting the item after being
found by index. Removing an item from a Max-Heap by index is not natively
supported, but uses the same technique as removing the root: swap the item with
the last leaf (the last item in the array), remove the item, and then swap the moved
item up or down the tree to restore the Max-Heap property.
Problem 5. Robot Wrangling
Dr. Squid has built a robotic arm from n+1 rigid bars called links, each connected
to the one before it with a rotating joint (n joints in total). Following standard
practice in robotics3, the orientation of each link is specified locally relative to the
orientation of the previous link. In mathematical notation, the change in
orientation at a joint can be specified using a 4 × 4 transformation matrix. Let M =
(M0, . . . , Mn1) be an array of transformation matrices associated with the arm,
where matrix Mk is the change in orientation at joint k, between links k and k + 1.

To compute the position of the end effector4, Dr. Squid will need the arm’s Q full
transformation: n-1 the ordered matrix product of the arm’s transformation
matrices, k=0 Mk = M0 ·M1 · . . . ·Mn-1. Assume Dr. Squid has a function matrix
multiply(M1,M2) that returns the matrix product5 M1 × M2 of any two 4 × 4
transformation matrices in O(1) time. While tinkering with the arm changing one
joint at a time, Dr. Squid will need to re-evaluate this matrix product quickly.
Describe a database to support the following worst-case operations to accelerate
Dr. Squid’s workflow:
Solution:
Store the matrices in a Sequence AVL tree T, where every node v stores a matrix
v.M and is augmented with v.P: the ordered product of all matrices in v’s subtree.
This property at a node v can be computed in O(1) time from the augmentations
of its children. Specifically, let PL
initialize(M) Initialize from an initial input configuration M in O(n)
time.
update joint(k,M) Replace joint k’s matrix Mk with matrix M in O(log n)
time.
full transformation() Return the arm’s current full transformation in O(1)
time.

and PR be v.left.P and v.right.P respectively (or the 4 × 4 identity matrix if the
respective left or right child is missing); then v.P = PL · v.M · PR which can be
computed in O(1) time using the provided matrix multiply function, so this
augmentation can be maintained. (Note that, because the number of items and
traversal order never changes, AVL behavior is not needed here, since no
operation will change the structure of the tree. So a static binary tree or even an
implicitly represented complete binary tree stored in an array, as in a binary
heap, would suffice.)
Now we support the operations:
• initialize(M): build T from the matrices in worst-case M in O(|M|) = O(n) time,
maintaining the new augmentation from the leaves to the root.
• update joint(k, M): find the node v containing matrix k in the traversal order
using get at(k) at the root of T in O(log n) time, and replace v.M with M. Then
recompute the augmentations up the tree in O(log n) time.
• full transformation(): the augmentation stored at the root of T corresponds
exactly to the arm’s full transformation, so simply return T.root.P in O(1) time.
Problem 6. πz2a Optimization

Liza Pover has found a Monominos pizza left over from some big-TEX recruiting
event. The pizza is a disc6 with radius z, having n toppings labeled 0, . . . , n � 1.
Assume z fits in a single machine word, so integer arithmetic on O(1) such
integers can be done in O(1) time. Each topping i:
• is located at Cartesian coordinates (xi, yi) where xi, yi are integers from range R
= {�z, . . . , z} (you may assume that all coordinates are distinct), and
• has integer tastiness ti ∈ R (note, topping tastiness can be negative, e.g., if it’s
pineapple7).
Liza wants to pick a point (x0 , y0 ) and make a pair of cuts from that point, one
going straight down and one going straight left, and take the resulting slice, i.e.,
the intersection of the pizza with the two half-planes x ≤ x0 and y ≤ y0 . The
tastiness of this slice is the sum of all ti such that xi ≤ x0 and yi ≤ y0 . Liza wants
to find a tastiest slice, that is, a slice of maximum tastiness. Assume there exists a
slice with positive tastiness.
(a) If point (x0 , y0 ) results in a slice with tastiness t 6= 0, show there exists i, j ∈
{0, 1, . . . , n � 1} such that point (xi, yj ) results in a slice of equal tastiness t
(i.e., a tastiest slice exists resulting from a point that is both vertically and
horizontally aligned with toppings).

Solution:
Pick the largest xi ≤ x0 and the largest yj ≤ y0 . Because t 6= 0, the slice contains
at least one topping, so there exists such toppings i and j. And since the slice S
corresponding to point (xi, yj ) contains exactly the same toppings as the slice
corresponding to (x0 , y0 ), then slice S also has tastiness t.
(b) To make finding a tastiest slice easier, show how to modify a Set AVL Tree so
that:
• it stores key–value items, where each item x contains a value x.val (in addition
to its key x.key on which the Set AVL is ordered);
• it supports a new tree-level operation max prefix() which returns in worstcase
O(1) time a pair (k∗, prefix(k∗)), where k∗ P is any key stored in the tree T that
maximizes the prefix sum, prefix(k) = {x.val | x ∈ T and x.key ≤ k} (that is, the
sum of all values of items whose keys are ≤ k); and
• all other Set AVL Tree operations maintain their running times.
Solution:
We augment the Set AVL Tree so that each node v stores three additional subtree
properties: • v.sum: the sum of all item values stored in v’s subtree, which can be

computed in O(1) time by: v.sum = v.left.sum + v.item.val + v.right.sum, or with
zeros for the left and right sums if the respective children do not exist. • v.max
prefix: max{prefix(k) | k ∈ subtree(v)}. We can compute the max prefix in the
subtree in O(1) time by comparing three values:
v.max_prefix = max(
v.left.max_prefix, # left
v.left.sum + v.item.val, #
middle
v.left.sum + v.item.val + v.right.max_prefix) #
right
where augmentations are considered zero on non-existent nodes.
• v.max prefix key: arg max{prefix(k) | k ∈ subtree(v)}. We can compute the
maximizing key in O(1) time based on which of the three cases is maximum in the
above computation: key is v.left.max prefix key if the left is maximizing,
v.item.key if the middle is maximizing, and v.right.max prefix key if the right is
maximizing.
Because these augmentations can be computed locally in O(1) time, they can be in
O(1) time via the augmentations stored at the root.

maintained without effecting the running times of the normal Set AVL Tree
operations.
support T.max prefix(), simply return
(T.root.max prefix, T.root.max prefix key)
(c) Using the data structure from part (b) as a black box, describe a worst-case
O(n log n)-time algorithm to return a triple (x, y,t), where point (x, y)
corresponds to a slice of maximum tastiness t.
Solution:
Sort the input topping points by their x coordinates in O(n log n) time (e.g., using
merge sort), and initialize the data structure T from part (b), initially empty. Then
for each topping (xi, yi, ti), insert it into T as a key-value item with key yi and
value ti in O(log n) time, and then evaluate the max prefix (y∗ , t∗ ) in T . The max
prefix t ∗ is then by definition the maximum tastiness of any slice with x
coordinate xi, specifically the slice corresponding to the point (xi, y∗ ). By
repeating this procedure for each topping sorted by x, we can compute the
maximum tastiness of any slice at xi for every xi in O(n log n) time (along with its
associated point). Since some slice of maximum tastiness exists with an x

coordinate at some xi for i ∈ {0, . . . , n � 1}, as argued in part (a), then taking the
maximum of all slices found in O(n) time will correctly return a tastiest slice
possible.
(ci) Write a Python function tastiest slice(toppings) that implements your
algorithm from part (c), including an implementation of your data structure from
part (b). You can download a code template containing some test cases from the
website.
Solution:
1 class Part_B_Node(BST_Node):
2 def subtree_update(A):
3 super().subtree_update()
4 A.sum = A.item.val # sum
5 if A.left: A.sum += A.left.sum
6 if A.right: A.sum += A.right.sum
7 left = -float(’inf’) # max prefix
8 right = -float(’inf’)
9 middle = A.item.val
10 if A.left:

11 left = A.left.max_prefix
12 middle += A.left.sum
13 if A.right:
14 right = middle + A.right.max_prefix
15 A.max_prefix = max(left, middle, right)
16 if left == A.max_prefix: # max prefix key
17 A.max_prefix_key = A.left.max_prefix_key
18 elif middle == A.max_prefix:
19 A.max_prefix_key = A.item.key
20 else:
21 A.max_prefix_key = A.right.max_prefix_key
22
23 class Part_B_Tree(Set_AVL_Tree):
24 def __init__(self):
25 super().__init__(Part_B_Node)
26
27 def max_prefix(self):
28 k = self.root.max_prefix_key
29 s = self.root.max_prefix
30 return (k, s)
31

32 def tastiest_slice(toppings):
33 B = Part_B_Tree() # use data structure from part (b)
34 X, Y, T = 0, 0, 0
35 n = len(toppings)
36 toppings.sort(key = lambda topping: topping[0])
37 for (x, y, t) in toppings:
38 B.insert(Key_Val_Item(y, t))
39 (Y_, T_) = B.max_prefix()
40 if T < T_:
41 X, Y, T = x, Y_, T_
42 return (X, Y, T)

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