Hash Tables, Heaps, and Priority Structures · Hashing

Hash Table

Hash tables support fast average-case lookup by turning keys into array positions and resolving collisions carefully.

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Student Focus

We connect hash tables to everyday map and set use before discussing implementation internals.

Guided Lesson Notes

Understanding Hash Table

Hash Table focuses on fast lookup, priority selection, and the difference between average-case access and ordered removal. Hash tables support fast average-case lookup by turning keys into array positions and resolving collisions carefully.

Model: separate the key or priority from the stored value; the structure exists to answer one operation very quickly.

Invariant: a hash table must send equal keys to the same place, while a heap must keep every parent no worse than its children.

Implementation: maps, sets, counters, priority queues, comparators, heapify steps, and careful handling of collisions or ties.

Use Hash Table when the problem asks for frequencies, duplicates, top-k values, smallest available item, repeated minimum merge, or dynamic priorities.

Visual Model

A small picture for Hash Table

Hash Table View

0: lime

1: empty

2: pear -> plum

3: apple

Heap View

2479121520

Key Ideas

Hash functions and equality
Collision strategies
Load factor and resizing

Practice Prompts

Build a frequency map for words in text.
Trace separate chaining after several insertions.

Vocabulary

Terms students should be able to say clearly

Key

The value used to find or group data in a hash table.

Collision

Two keys landing in the same hash-table location.

Load factor

How full a hash table is relative to its capacity.

Priority

The ordering rule used by a priority queue or heap.

Heap property

The parent-child ordering rule maintained by a heap.

Amortized cost

Average cost over many operations, including occasional expensive cleanup.

Worked Example

Worked example: tracing Hash Table

Trace a small input, focusing on hash functions and equality.

1Write a small input where Hash Table is clearly useful.
2Label the part of the input related to Hash functions and equality.
3Perform one operation and explain which invariant is still true afterward.
4Run a second operation that touches an edge case, such as an empty side, duplicate value, boundary index, disconnected vertex, or tie.
5Finish by saying which operation dominates the runtime and why.

Complexity Check

Costs students should be able to explain

Operation	Typical Cost	Reason
Hash lookup	Average O(1)	Good hashing makes key lookup nearly constant on typical inputs.
Heap insert/remove	O(log n)	A heap restores its invariant along one root-to-leaf path.
Build from all values	O(n) to O(n log n)	The cost depends on whether heapify or repeated insertion is used.

Common Mistakes

What to watch while practicing

Coding before drawing the structure or state changes.
Forgetting the invariant that makes the algorithm correct.
Testing only the sample input and skipping boundary cases.
Giving Big-O without explaining which operation dominates the work.

Interview-Style Coding Problem

Hash Table interview problem: Longest Consecutive Sequence

Interview medium

Problem

Given an unsorted array, find the length of the longest run of consecutive integer values.

Input

The first line contains n. The second line contains n integers.

Output

Print the length of the longest consecutive run.

Sample Input

6
100 4 200 1 3 2

Sample Output

Why the sample works

The values 1, 2, 3, and 4 form the longest consecutive run.

Approach

1Put all values into a hash set.
2Only start counting from a value if value - 1 is not present.
3Walk forward while consecutive values exist.
4Each value is part of at most one walk.

O(n) expected time and O(n) extra space.

Java Solution

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    Set<Integer> values = new HashSet<>();
    for (int i = 0; i < n; i++) values.add(sc.nextInt());

    int best = 0;
    for (int x : values) {
      if (!values.contains(x - 1)) {
        int length = 1;
        while (values.contains(x + length)) {
          length++;
        }
        best = Math.max(best, length);
      }
    }

    System.out.println(best);
  }
}

Python Solution

n = int(input())
values = set(map(int, input().split()))

best = 0
for x in values:
    if x - 1 not in values:
        length = 1
        while x + length in values:
            length += 1
        best = max(best, length)

print(best)

Practice Challenge

Make the idea your own

Create a two-minute explanation of Hash Table: define it, trace one example, name one edge case, and give the runtime of the main operation.

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