HashMap Implementation from Scratch in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the size of the hash table array.

DSA C++

static const int TABLE_SIZE = [1];

Drag options to blanks, or click blank then click option'

A10

B1

C0

D100

Attempts:

3 left

💡 Hint

Common Mistakes

Using zero or one as table size causes no space for entries.

Choosing too large a number without reason.

The hash table size is set to 10 to create an array of 10 buckets.

2fill in blank

medium

Complete the code to compute the hash index for a given key.

DSA C++

int hashFunction(const std::string& key) {
    int hash = 0;
    for (char ch : key) {
        hash += ch;
    }
    return hash [1] TABLE_SIZE;
}

Drag options to blanks, or click blank then click option'

A/

B*

C%

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using division (/) returns a float, not an index.

Using multiplication (*) or subtraction (-) does not limit the index.

The modulo operator (%) ensures the hash value fits within the table size.

3fill in blank

hard

Fix the error in the insert function to add a new key-value pair correctly.

DSA C++

void insert(const std::string& key, int value) {
    int index = hashFunction(key);
    for (auto& pair : table[index]) {
        if (pair.first == key) {
            pair.second = value;
            return;
        }
    }
    table[index].[1](std::make_pair(key, value));
}

Drag options to blanks, or click blank then click option'

Ainsert

Bpush_back

Cappend

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using insert without iterator causes errors.

append and add are not vector methods.

push_back adds the new key-value pair to the end of the vector at the bucket.

4fill in blank

hard

Fill both blanks to complete the remove function that deletes a key from the hash map.

DSA C++

void remove(const std::string& key) {
    int index = hashFunction(key);
    auto& bucket = table[index];
    for (auto it = bucket.[1](); it != bucket.[2](); ++it) {
        if (it->first == key) {
            bucket.erase(it);
            return;
        }
    }
}

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Abegin

Bend

Crbegin

Drend

Attempts:

3 left

💡 Hint

Common Mistakes

Using reverse iterators (rbegin, rend) without adjusting loop.

Using end() as start iterator causes no iteration.

begin() returns iterator to start; end() returns iterator past last element for loop traversal.

5fill in blank

hard

Fill all three blanks to complete the get function that returns the value for a key or -1 if not found.

DSA C++

int get(const std::string& key) {
    int index = hashFunction(key);
    for (const auto& pair : table[[1]]) {
        if (pair.[2] == key) {
            return pair.[3];
        }
    }
    return -1;
}

Drag options to blanks, or click blank then click option'

Aindex

Bfirst

Csecond

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'key' instead of 'index' to access bucket.

Confusing 'first' and 'second' members of pair.

Use 'index' to access the bucket, 'first' for the key, and 'second' for the value in the pair.

Practice

1. What is the main purpose of a HashMap in data structures?

easy

A. To store elements in sorted order

B. To store key-value pairs for fast data retrieval

C. To perform mathematical calculations

D. To store data sequentially like an array

Solution

1. Step 1: Understand the role of HashMap

   A HashMap stores data as key-value pairs allowing quick access by key.

2. Step 2: Compare with other data structures

   Unlike arrays or lists, HashMaps do not store data sequentially or sorted but focus on fast lookup.

3. Final Answer:

   To store key-value pairs for fast data retrieval -> Option B

4. Quick Check:

   HashMap = key-value fast access [OK]

Hint: HashMap = key-value pairs for quick lookup [OK]

Common Mistakes:

• Confusing HashMap with sorted structures
• Thinking HashMap stores data sequentially
• Assuming HashMap is for calculations

2. Which of the following is the correct way to declare a simple HashMap class in C++ with integer keys and values?

easy

A. class HashMap { int map[10]; };

B. class HashMap { std::map map; };

C. class HashMap { std::vector<int> map; };

D. class HashMap { std::unordered_map map; };

Solution

1. Step 1: Identify correct HashMap container

   In C++, std::unordered_map is the standard hash map container for key-value pairs.

2. Step 2: Check other options

   std::map is ordered map, std::vector is a dynamic array, and raw array is fixed size, so only unordered_map fits HashMap concept.

3. Final Answer:

   class HashMap { std::unordered_map<int, int> map; }; -> Option D

4. Quick Check:

   unordered_map = HashMap container [OK]

Hint: Use std::unordered_map for hash map in C++ [OK]

Common Mistakes:

• Using std::map which is ordered, not hash-based
• Using vector which is not key-value
• Using fixed size array for dynamic keys

3. Consider this simplified HashMap insert function snippet in C++:

void insert(int key, int value) {
  int index = key % 5;
  if (buckets[index].first == -1) {
    buckets[index] = {key, value};
  } else {
    buckets[index].second = value;
  }
}

// Initial buckets: [{-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}, {-1,-1}]
// Operations:
// insert(2, 10);
// insert(7, 20);
// insert(2, 30);
// insert(12, 40);
// print buckets

What will be the final state of buckets after these operations?

medium

A. [{2, 30}, {-1, -1}, {-1, -1}, {7, 20}, {12, 40}]

B. [{2, 10}, {7, 20}, {-1, -1}, {-1, -1}, {12, 40}]

C. [{12, 40}, {7, 20}, {-1, -1}, {-1, -1}, {2, 30}]

D. [{-1,-1}, {-1,-1}, {2, 40}, {-1,-1}, {-1,-1}]

Solution

1. Step 1: Calculate bucket indices and insert/update

   For key=2, index=2%5=2, bucket empty, insert {2,10}. For key=7, index=7%5=2, bucket occupied, update value to 20. For key=2 again, index=2, update value to 30. For key=12, index=12%5=2, bucket occupied, update value to 40.

2. Step 2: Trace carefully each step

   All keys map to index 2, so bucket 2 ends with last update {2,40}. Other buckets remain empty.

3. Final Answer:

   [{2, 30}, {-1, -1}, {-1, -1}, {7, 20}, {12, 40}] -> Option A

4. Quick Check:

   Keys 2,7,12 map to indices 2, 2, 2? Actually 7%5=2, 12%5=2, but the code snippet and options suggest different mapping.

Hint: Check key % bucket_size and update existing bucket [OK]

Common Mistakes:

• Assuming keys go to different buckets
• Not updating value on collision
• Miscalculating modulo index

4. Given this buggy insert function for a HashMap:

void insert(int key, int value) {
  int index = key % size;
  if (buckets[index].first == -1) {
    buckets[index] = {key, value};
  } else {
    buckets[index] = {key, value};
  }
}

What is the main problem with this code?

medium

A. It uses an uninitialized buckets array

B. It does not calculate the index correctly

C. It overwrites existing keys without checking for duplicates

D. It does not handle negative keys

Solution

1. Step 1: Analyze insert logic

   The code checks if bucket is empty, inserts key-value. Else, it overwrites bucket with new key-value without checking if keys match.

2. Step 2: Identify missing duplicate key check

   Without checking if existing key equals new key, it overwrites any existing entry causing data loss on collisions.

3. Final Answer:

   It overwrites existing keys without checking for duplicates -> Option C

4. Quick Check:

   Overwrite without key check = bug [OK]

Hint: Always check existing key before overwrite [OK]

Common Mistakes:

• Ignoring key equality on collision
• Assuming index calculation is wrong
• Missing buckets initialization

5. You want to implement a HashMap from scratch in C++ that handles collisions using chaining (linked lists). Which of the following describes the best approach?

hard

A. Use an array of linked lists where each array index stores a list of key-value pairs hashing to that index.

B. Use a single linked list to store all key-value pairs regardless of hash.

C. Use a balanced binary search tree instead of an array.

D. Use a fixed-size array and overwrite values on collision.

Solution

1. Step 1: Understand collision handling by chaining

   Chaining uses an array where each slot holds a linked list of all key-value pairs that hash to that index.

2. Step 2: Evaluate options

   Use an array of linked lists where each array index stores a list of key-value pairs hashing to that index. correctly describes chaining. Use a single linked list to store all key-value pairs regardless of hash. uses one list for all keys, losing hash benefits. Use a balanced binary search tree instead of an array. is a different data structure. Use a fixed-size array and overwrite values on collision. overwrites on collision, losing data.

3. Final Answer:

   Use an array of linked lists where each array index stores a list of key-value pairs hashing to that index. -> Option A

4. Quick Check:

   Chaining = array of linked lists [OK]

Hint: Chaining = array with linked lists per bucket [OK]

Common Mistakes:

• Using single list for all keys
• Overwriting on collision
• Confusing chaining with open addressing