Hash Tables and Maps, Lecture notes of Data Structures and Algorithms

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Hash Tables

15 - 121 Fall 2020

Margaret Reid-Miller

Today

• Sets and Maps review

• Hash Tables

• Next time

• hashCodes

• Priority Queues

Set

(sometimes called a bag) A set is “bag” of objects

• No duplicates with respect to .equals()
• Membership Operations I want to be fast:
  • Does the set contain this element?
  • Add this element to the set
• Remove an element from the set Tom Sue Sally Fred Dave

Map

(also called dictionary or associative array)

A map is a table of (key,value) pairs.

• Indexed by key (must be unique).

• Many keys can “map” to the same value.

Operations I want to be fast:

• Get Tom’s section

• Set Dave’s section to B

• Remove Fred from the class

Name Section Tom A Fred B Dave A Sally C

Both TreeSet & TreeMap use a balanced binary search tree called a “red-black tree". Red-Black balanced binary search trees:

• The height of a red-black tree is guaranteed to be 2 log n.
• Every time you add or remove an element, the tree may be restructured to maintain balance.
• The runtime to rebalance the tree is worst case O(log n).

Thus

• Operations contains/add/remove for Sets and get/set/remove for map have worst-case O(log n) runtime.

Worst case O(log n) time! Great! We’re done! The course is over! Yay! Unless…

Really Big Array (?)

Use the student id as the index into a really big array: contains: O(1), add: O(1), remove: O(1). Yay!!! Problem: 0 1 … 151212018 à id: 151- 21 - 2018 name: Dave … 999999999 Memory hog: The range of student id values is independent on the number of students (size of the set).

Moderate Size Array (better)

Key Idea: Use the key to compute an index into a moderate size array.

• Want: contains, add, remove: O(1), memory: O(n) Example : Use last two digits of the student id
• Problem: 0 1 18 à (^) id: 151 - 21 - 2020 name: Dave 99 Two or more students might have the same last two digits.

Collisions

• Can we prevent collisions when we don’t know in advance which keys will be used in the set? - No. Since the number of possible keys is much greater than the size of the hash table, there must be two keys that map to the same index. - Any set that contains those two keys will have a collision.
• Pigeonhole Principle : If you put more than n items into n bins, then at least one bin contains more than one item. all possible keys S S

The Birthday Paradox

• How likely are two keys going to hash to the same index?

• Probability that none of n people have the same

birthday:

p’ = 1(364/365)(363/365)…((365- n +1)/365)

• Probability at least two people have the same birthday is p = 1 – p’ When n = 23, p = 0.5. When n = 30, p = 0. When n = 50, p = 0.97 !! Surprisingly likely!

Handling Collisions

1. Open Addressing (topic for 15 - 451) 2. Separate chaining – Each index of array contains all the elements that hash to that index (called a bucket ) What data structure should we use to maintain a bucket?

• Often a linked list because:
  • Buckets are small (few collisions)
  • Linked lists easy to implement
  • Many buckets can be empty and empty linked lists take no storage
  • No additional constraints such as Comparable

Separate Chaining using a linked list 10 … Dave … Tom … Sue 65 null null null null null

Runtime

What is the worst-case runtime for contains, add, remove? O(n) – all the keys hash to the same index What is the best-case runtime? O(1) – only a few keys map to any one index What is the expected runtime? O(1) – assuming the hash function is good, and the hash table is not too full

Load Factor

Load Factor: (number of elements) / (length of array) What is the expected size of a bucket? The load factor What is a good load factor? A small constant so that the linked list stay short, even the longest ones. Java uses a default value of 0. Can the load factor be larger than 1? Yes