CS 253: Algorithms
Chapter 6
HeapsortAppendix B.5
Credit: Dr. George Bebis
2
Special Types of Trees
Def: Full binary tree a binary tree in which each node is either a leaf or has degree exactly 2.
Def: Complete binary treea binary tree in which all leaves are on the same level and all internal nodes have degree 2.
Full binary tree
214 8
1
167
4
3
9 1012
Complete binary tree
2
1
16
4
3
9 10
Definitions Height of a node = the number of edges on the
longest simple path from the node down to a leaf
Level of a node = the length of a path from the root to the node
Height of tree = height of root node
214 8
116
4
39 10
Height of root = 3
Height of (2)= 1 Level of (10)= 2
Useful Properties
214 8
116
4
39 10
Height of root = 3
Height of (2)= 1 Level of (10)= 2
• There are at most 2l nodes at level (or depth) l of a binary tree
• A binary tree of height h has at most (2h+1 – 1) nodes
• **here A binary tree with n nodes has height at least lgn. (see Ex 6.1-1&2)
1212
122 11
0
h
hh
i
in
5
The Heap Data StructureDef: A heap is a nearly complete binary tree
with the following two properties:◦ Structural property: all levels are full, except
possibly the last one, which is filled from left to right◦ Order (heap) property: for any node x,
Parent(x) ≥ x
Heap
57
8
42
From the heap property, it follows that:
The root is the maximum element of the heap!
Array Representation of Heaps
A heap can be stored as an array A.
◦ Root of tree is A[1]◦ Left child of A[i] = A[2i]◦ Right child of A[i] = A[2i
+ 1]◦ Parent of A[i] = A[ i/2 ]◦ Heapsize[A] ≤ length[A]
The elements in the subarray A[(n/2+1) .. n] are leaves
7
Heap Types
Max-heaps (largest element at root), have the max-heap property: ◦ for all nodes i, excluding the root: A[PARENT(i)] ≥ A[i]
Min-heaps (smallest element at root), have the min-heap property:◦ for all nodes i, excluding the root: A[PARENT(i)] ≤ A[i]
8
Operations on HeapsMaintain/Restore the max-heap property
◦ MAX-HEAPIFY
Create a max-heap from an unordered array◦ BUILD-MAX-HEAP
Sort an array in place◦ HEAPSORT
Priority queues
9
Maintaining the Heap PropertyMAX-HEAPIFY
Suppose a node i is smaller than a child and Left and Right subtrees of i are max-
heaps
How do you fix it?
To eliminate the violation: Exchange node i with the larger child Move down the tree Continue until node is not smaller than
children
◦ MAX-HEAPIFY
ExampleMAX-HEAPIFY(A, 2, 10)
A[2] violates the heap property
A[2] A[4]
A[4] violates the heap property
A[4] A[9]
Heap property restored
Maintaining the Heap Property
Assumptions:◦Left and Right
subtrees of i are max-heaps
◦A[i] may be smaller than its children
Alg: MAX-HEAPIFY(A, i, n)1. l ← LEFT(i)2. r ← RIGHT(i)3. if l ≤ n and A[l] > A[i]4. then largest ←l5. else largest ←i6. if r ≤ n and A[r] > A[largest]7. then largest ←r8. if largest i9. then exchange A[i] ↔
A[largest]10. MAX-HEAPIFY(A,
largest, n)
MAX-HEAPIFY Running Time It traces a path from the root to a leaf (longest path
length h)
At each level, it makes two comparisons
Total number of comparisons = 2h
Running Time of MAX-HEAPIFY = O(2h) = O(lgn)
13
Building a Heap
Alg: BUILD-MAX-HEAP(A)1. n = length[A]2. for i ← n/2 downto 13. do MAX-HEAPIFY(A, i, n)
Convert an array A[1 … n] into a max-heap (n = length[A])The elements in the subarray A[(n/2+1) .. n] are leavesApply MAX-HEAPIFY on elements between 1 and n/2 (bottom up)
214 8
116
7
4
39 10
1
2 3
4 5 6 7
8 9 10
4 1 3 2 16 9 10 14 8 7A:
Example: A4 1 3 2 16 9 10 14 8 7
214 8
116
7
4
39 10
1
2 3
4 5 6 7
8 9 10
142 8
116
7
4
109 3
1
2 3
4 5 6 7
8 9 10
214 8
116
7
4
39 10
1
2 3
4 5 6 7
8 9 10 142 8
116
7
4
39 10
1
2 3
4 5 6 7
8 9 10
142 8
167
1
4
109 3
1
2 3
4 5 6 7
8 9 10 82 4
147
1
16
109 3
1
2 3
4 5 6 7
8 9 10
i = 5 i = 4 i = 3
i = 2 i = 1
15
Running Time of BUILD MAX HEAP
Running time: O(nlgn)This is not an asymptotically tight upper bound! Why?
Alg: BUILD-MAX-HEAP(A)
1. n = length[A]
2. for i ← n/2 downto 1
3. do MAX-HEAPIFY(A, i, n)
O(lgn) O(n)
Running Time of BUILD MAX HEAP MAX-HEAPIFY takes O(h) the cost of MAX-HEAPIFY on a
node i proportional to the height of the node i in the tree
Height Levelh0 = 3 (lgn)
h1 = 2
h2 = 1
h3 = 0
i = 0
i = 1
i = 2
i = 3 (lgn)
No. of nodes20
21
22
23
hi = h – i height of the heap rooted at level ini = 2i number of nodes at level i
)(2)(00
nOihhnnTh
i
ii
h
ii
see the next slide
i
h
iihnnT
0
)( Cost of HEAPIFY at level i (# of nodes at that level)
ihh
i
i 0
2 Replace the values of ni and hi computed before
hh
iih
ih 220
Multiply by 2h both at the nominator and denominator
and write 2i as (1/2-i)
h
kk
h k0 2
2 Change variables: k = h - i
0 2k
k
kn The sum above is smaller than the sum of all elements to and h = lgn
)(nO Running time of BUILD-MAX-HEAP
T(n) = O(n)
Running Time of BUILD MAX HEAP
2)2/11(
2/1
n x=1/2 in the summation
20 )1( x
xkxk
k
18
Goal:◦ Sort an array using heap representations
Idea:◦ Build a max-heap from the array◦ Swap the root (the maximum element) with the
last element in the array◦ “Discard” this last node by decreasing the heap
size◦ Call MAX-HEAPIFY on the new root◦ Repeat this process until only one node remains
Heapsort
19
Example: A=[7, 4, 3, 1, 2]
MAX-HEAPIFY(A, 1, 4) MAX-HEAPIFY(A, 1, 3) MAX-HEAPIFY(A, 1, 2)
MAX-HEAPIFY(A, 1, 1)
20
Alg: HEAPSORT(A)
1. BUILD-MAX-HEAP(A)2. for i ← length[A] downto 23. do exchange A[1] ↔ A[i]4. MAX-HEAPIFY(A, 1, i - 1)
Running time: O(nlgn) Can be shown to be Θ(nlgn)
O(n)
O(lgn)
n-1 times
Priority QueuesEach element is associated with a value (priority)The key with the highest (lowest) priority is extracted first
Support the following operations:◦ INSERT(S, x) : inserts element x into set S◦ EXTRACT-MAX(S) : removes and returns element of S with largest
key◦ MAXIMUM(S) : returns element of S with largest key◦ INCREASE-KEY(S, x, k) : increases value of element x’s key to k
(Assume k ≥ x’s current key value)
14 5
22
HEAP-MAXIMUMGoal:
◦ Return the largest element of the heap
Alg: HEAP-MAXIMUM(A)1. return A[1]
Running time: O(1)
Heap A:
Heap-Maximum(A) returns 7
23
HEAP-EXTRACT-MAXGoal:
◦ Extract the largest element of the heap (i.e., return the max value and also remove that element from the heap
Idea: ◦ Exchange the root element with the last◦ Decrease the size of the heap by 1 element◦ Call MAX-HEAPIFY on the new root, on a heap of
size n-1Heap A: Root is the largest element
Example: HEAP-EXTRACT-MAX
82 4
147
1
16
109 3
max = 168
2 4
147
1
109 3
Heap size decreased with 1
42 1
87
14
109 3
Call MAX-HEAPIFY(A, 1, n-1)
25
HEAP-EXTRACT-MAX
Alg: HEAP-EXTRACT-MAX(A, n)1. if n < 1
2. then error “heap underflow”3. max ← A[1]4. A[1] ← A[n]5. MAX-HEAPIFY(A, 1, n-1) % remakes heap6. return max
Running time: O(lgn)
HEAP-INCREASE-KEYGoal:
◦ Increases the key of an element i in the heap
Idea:◦ Increment the key of A[i] to its new value◦ If the max-heap property does not hold
anymore: traverse a path toward the root to find the proper place for the newly increased key
82 4
147
1
16
109 3i
Key [i] ← 15
Example: HEAP-INCREASE-KEY
142 8
157
1
16
109 3
i
82 4
147
1
16
109 3i
Key [i ] ← 15
82 15
147
1
16
109 3i
152 8
147
1
16
109 3
i
Alg: HEAP-INCREASE-KEY(A, i, key)
1. if key < A[i]2. then error “new key is smaller than current
key”3. A[i] ← key4. while i > 1 and A[PARENT(i)] < A[i]5. do exchange A[i] ↔ A[PARENT(i)]6. i ← PARENT(i)
Running time: O(lgn)
82 4
147
1
16
109 3i
Key [i] ← 15
29
-
MAX-HEAP-INSERTGoal:
Inserts a new element into a max-heap
Idea: Expand the max-heap with a
new element whose key is -
Calls HEAP-INCREASE-KEY to set the key of the new node to its correct value and maintain the max-heap property
82 4
147
1
16
109 3
158
2 4
147
1
16
109 3
30
Example: MAX-HEAP-INSERT
-8
2 4
147
1
16
109 3
Insert value 15:Start by inserting -
158
2 4
147
1
16
109 3
Increase the key to 15Call HEAP-INCREASE-KEY on A[11] = 15
78
2 4
1415
1
16
109 3
The restored heap containingthe newly added element
78
2 4
1514
1
16
109 3
Alg: MAX-HEAP-INSERT(A, key, n)
1. heap-size[A] ← n + 12. A[n + 1] ← - 3. HEAP-INCREASE-KEY(A, n + 1, key)
Running time: O(lgn)
-8
2 4
147
1
16
109 3
32
SummaryWe can perform the following operations on
heaps:
◦ MAX-HEAPIFY O(lgn)◦ BUILD-MAX-HEAP O(n)◦ HEAP-SORT O(nlgn)◦ MAX-HEAP-INSERT O(lgn)◦ HEAP-EXTRACT-MAX O(lgn)◦ HEAP-INCREASE-KEY O(lgn)◦ HEAP-MAXIMUM O(1)
Average:
O(lgn)
33
Priority Queue Using Linked List
Average: O(n)
Increase key: O(n)
Extract max key: O(1)
12 4
Problemsa. What is the maximum number of
nodes in a max heap of height h?
b.What is the maximum number of leaves?
c. What is the maximum number of internal nodes?
d.Assuming the data in a max-heap are distinct, what are the possible locations of the second-largest element?
35
Demonstrate, step by step, the operation of Build-Heap on the array A = [5, 3, 17, 10, 84, 19, 6, 22, 9]
Let A be a heap of size n. Give the most efficient algorithm for the following tasks:
(a) Find the sum of all elements(b) Find the sum of the largest lgn elements
Problems
Recommended
