BST vs Hash Map Trade-offs for Ordered Data in DSA C - Compare & Choose

Challenge - 5 Problems

🧠 Conceptual

intermediate

Why choose a BST over a Hash Map for ordered data?

Which of the following is the main reason to use a Binary Search Tree (BST) instead of a Hash Map when you need to keep data ordered?

A BSTs maintain elements in sorted order, allowing in-order traversal to get sorted data easily.

B BSTs have faster average lookup time than Hash Maps.

C BSTs use less memory than Hash Maps in all cases.

D BSTs guarantee constant time insertion and deletion.

💡 Hint

Think about how data is stored and accessed in both structures.

BSTs keep data sorted by their nature, so you can get ordered data by traversing them in order. Hash Maps do not keep order.

❓ Predict Output

intermediate

Output of in-order traversal on BST vs Hash Map keys

Given the following data inserted in this order: 5, 3, 7, 2, 4, 6, 8, what is the output of an in-order traversal of a BST and the keys iteration of a Hash Map?

DSA C

BST insertion order: 5, 3, 7, 2, 4, 6, 8
Hash Map keys: 5, 3, 7, 2, 4, 6, 8 (insertion order)

Output BST in-order traversal:

Output Hash Map keys iteration:

A

BST in-order: 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null
Hash Map keys: 2, 3, 4, 5, 6, 7, 8

B

BST in-order: 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null
Hash Map keys: 5, 3, 7, 2, 4, 6, 8

C

BST in-order: 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> null
Hash Map keys: 5, 3, 7, 2, 4, 6, 8

D

BST in-order: 5 -> 3 -> 7 -> 2 -> 4 -> 6 -> 8 -> null
Hash Map keys: 2, 3, 4, 5, 6, 7, 8

💡 Hint

BST in-order traversal always returns sorted keys. Hash Map keys order depends on insertion or internal hashing.

BST in-order traversal visits nodes in ascending order. Hash Map keys iteration order is not guaranteed to be sorted and often follows insertion order or internal hashing.

🔧 Debug

advanced

Why does this Hash Map fail to maintain order?

Consider this code snippet that inserts keys 1 to 5 into a Hash Map and then prints keys expecting sorted order. Why does it not print keys in sorted order?

DSA C

hash_map = {}
for i in [3, 1, 4, 5, 2]:
    hash_map[i] = str(i)

for key in hash_map:
    print(key, end=' ')

A The keys are sorted but the print statement reverses the order.

B The keys are sorted internally but printing code is incorrect.

C Hash Maps automatically sort keys but only when keys are strings.

D Hash Maps do not maintain any order of keys; iteration order is arbitrary or insertion order depending on implementation.

💡 Hint

Think about how Hash Maps store keys and how iteration order works.

Hash Maps do not guarantee sorted order of keys. Iteration order depends on insertion or internal hashing, so keys appear unordered.

❓ Predict Output

advanced

Time complexity difference for range queries

Which data structure allows efficient range queries (e.g., find all keys between 10 and 20) and why?

DSA C

Data structures: BST and Hash Map
Operation: Find all keys between 10 and 20 inclusive

A BST allows efficient range queries by traversing only relevant nodes; Hash Map requires checking all keys.

B Hash Map allows efficient range queries using hashing; BST requires full traversal.

C Both BST and Hash Map have the same efficiency for range queries.

D Neither BST nor Hash Map can perform range queries efficiently.

💡 Hint

Think about how data is stored and accessed in each structure.

BSTs keep data sorted, so you can traverse only nodes in the range. Hash Maps have no order, so you must check all keys.

🚀 Application

expert

Choosing data structure for ordered data with frequent insertions and lookups

You need to store a large set of numeric keys with frequent insertions and lookups. You also need to retrieve keys in sorted order often. Which data structure is best and why?

A Use an unsorted array because insertions are fast and sorting can be done before lookups.

B Use a Hash Map because it has O(1) average insertions and lookups and sorting can be done when needed.

C Use a self-balancing BST because it keeps keys sorted and supports O(log n) insertions and lookups.

D Use a linked list because it maintains insertion order and supports fast lookups.

💡 Hint

Consider both the need for order and performance of operations.

Self-balancing BSTs keep keys sorted and provide efficient O(log n) insertions and lookups, making them suitable for frequent ordered retrievals.

Practice

1. Which data structure is best suited for maintaining data in sorted order with efficient range queries?

easy

A. Hash Map

B. Stack

C. Binary Search Tree (BST)

D. Queue

Solution

1. Step 1: Understand data ordering requirements

   Range queries and sorted data require maintaining order among elements.

2. Step 2: Compare BST and Hash Map for ordering

   BST keeps elements in sorted order by design; Hash Map does not maintain order.

3. Final Answer:

   Binary Search Tree (BST) -> Option C

4. Quick Check:

   Ordered data = BST [OK]

Hint: For sorted data and range queries, choose BST [OK]

Common Mistakes:

• Thinking hash maps maintain order
• Confusing stack or queue as ordered structures
• Ignoring the need for range queries

2. Which of the following is the correct way to declare a hash map in C using standard libraries?

easy

A. #include \nhash_map map;

B. #include \nstruct hashmap map;

C. #include \nint map[100];

D. #include \nGHashTable *map = g_hash_table_new(g_int_hash, g_int_equal);

Solution

1. Step 1: Identify standard C hash map usage

   C standard library does not have built-in hash maps; GLib provides GHashTable.

2. Step 2: Check options for valid syntax

   #include <glib.h>\nGHashTable *map = g_hash_table_new(g_int_hash, g_int_equal); uses GLib's GHashTable correctly; others are invalid or incomplete.

3. Final Answer:

   #include <glib.h>\nGHashTable *map = g_hash_table_new(g_int_hash, g_int_equal); -> Option D

4. Quick Check:

   GLib hash map syntax = #include <glib.h>\nGHashTable *map = g_hash_table_new(g_int_hash, g_int_equal); [OK]

Hint: Use GLib for hash maps in C, not standard library [OK]

Common Mistakes:

• Assuming C has built-in hash map
• Using C++ syntax in C code
• Confusing arrays with hash maps

3. Consider the following code snippet using a BST and a hash map for storing integer keys. What will be the output of the in-order traversal of the BST after inserting keys 5, 3, 7, 2, 4?

medium

A. 2 -> 3 -> 4 -> 5 -> 7 -> null

B. 5 -> 3 -> 7 -> 2 -> 4 -> null

C. 7 -> 5 -> 4 -> 3 -> 2 -> null

D. 3 -> 2 -> 4 -> 5 -> 7 -> null

Solution

1. Step 1: Insert keys into BST

   Insert 5 as root, then 3 left of 5, 7 right of 5, 2 left of 3, 4 right of 3.

2. Step 2: Perform in-order traversal

   In-order visits left subtree, root, right subtree: 2, 3, 4, 5, 7.

3. Final Answer:

   2 -> 3 -> 4 -> 5 -> 7 -> null -> Option A

4. Quick Check:

   In-order BST traversal = sorted keys [OK]

Hint: In-order traversal of BST gives sorted keys [OK]

Common Mistakes:

• Confusing insertion order with traversal order
• Mixing pre-order or post-order traversal
• Assuming hash map order applies to BST

4. The following C code attempts to insert keys into a BST but causes a runtime error. What is the main issue?

typedef struct Node {
  int key;
  struct Node *left, *right;
} Node;

Node* insert(Node* root, int key) {
  if (root == NULL) {
    Node* newNode = malloc(sizeof(Node));
    newNode->key = key;
    return newNode;
  }
  if (key < root->key)
    root->left = insert(root->left, key);
  else
    root->right = insert(root->right, key);
  return root;
}

medium

A. newNode's left and right pointers are not initialized

B. malloc is not included via stdlib.h

C. The function does not handle duplicate keys

D. The root pointer is not updated in main

Solution

1. Step 1: Check new node initialization

   newNode's left and right pointers are not set, so they contain garbage values.

2. Step 2: Understand impact on BST

   Uninitialized pointers cause undefined behavior or runtime errors during traversal or further insertions.

3. Final Answer:

   newNode's left and right pointers are not initialized -> Option A

4. Quick Check:

   Initialize pointers to NULL to avoid errors [OK]

Hint: Always initialize new node pointers to NULL [OK]

Common Mistakes:

• Ignoring pointer initialization
• Assuming malloc auto-initializes memory
• Confusing duplicate key handling with runtime error

5. You need to store a large dataset of user IDs with associated scores. You want to quickly find users with scores in a specific range and also retrieve scores by user ID efficiently. Which combination of data structures is best?

hard

A. Use only a hash map keyed by user ID

B. Use a hash map for user ID to score and a balanced BST keyed by score for range queries

C. Use only a BST keyed by user ID

D. Use an array sorted by user ID

Solution

1. Step 1: Identify requirements

   Fast lookup by user ID and efficient range queries by score are needed.

2. Step 2: Match data structures to needs

   Hash map gives O(1) user ID lookup; balanced BST keyed by score supports range queries efficiently.

3. Step 3: Combine for best performance

   Using both structures together meets both requirements effectively.

4. Final Answer:

   Use a hash map for user ID to score and a balanced BST keyed by score for range queries -> Option B

5. Quick Check:

   Hash map + BST combo = best for lookup + range [OK]

Hint: Combine hash map for lookup and BST for range queries [OK]

Common Mistakes:

• Using only one structure and missing efficiency
• Using array which is slow for lookups
• Ignoring need for range queries