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DSA Goprogramming~5 mins

Collision Handling Using Chaining in DSA Go - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is collision handling in hashing?
Collision handling is a method to resolve situations when two keys hash to the same index in a hash table.
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beginner
Explain chaining as a collision handling technique.
Chaining stores multiple elements at the same index using a linked list or similar structure, allowing all collided keys to be stored together.
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beginner
What data structure is commonly used in chaining to handle collisions?
A linked list is commonly used to store all elements that hash to the same index in chaining.
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intermediate
How does chaining affect the average time complexity of search in a hash table?
With chaining, average search time is O(1 + α), where α is the load factor (number of elements divided by number of buckets).
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intermediate
Why is chaining preferred over open addressing in some cases?
Chaining handles collisions gracefully even when the table is full, and it is easier to implement deletion of elements.
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What happens when two keys hash to the same index in chaining?
AOne key replaces the other
BThe second key is discarded
CThe hash table resizes immediately
DBoth keys are stored in a linked list at that index
Which data structure is typically used in chaining to store collided elements?
ALinked list
BArray
CStack
DQueue
What is the load factor in a hash table?
ANumber of elements divided by number of buckets
BNumber of collisions divided by number of elements
CNumber of buckets divided by number of elements
DNumber of keys hashed
How does chaining handle deletion of elements?
ADeletion is not possible in chaining
BBy removing the element from the linked list at the index
CBy rehashing the entire table
DBy marking the element as deleted but keeping it
What is the average search time complexity in chaining when load factor is low?
AO(n)
BO(log n)
CO(1)
DO(n log n)
Describe how chaining resolves collisions in a hash table.
Think about how multiple items can share the same bucket.
You got /3 concepts.
    Explain the impact of load factor on the performance of chaining.
    Consider how crowded the buckets get.
    You got /3 concepts.

      Practice

      (1/5)
      1. What is the main idea behind collision handling using chaining in hash tables?
      easy
      A. Use open addressing to find another empty slot
      B. Resize the hash table to avoid collisions
      C. Store multiple elements in a linked list at the same index when collisions occur
      D. Ignore collisions and overwrite existing data

      Solution

      1. Step 1: Understand collision in hash tables

        When two keys hash to the same index, a collision happens.
      2. Step 2: How chaining handles collisions

        Chaining stores all collided elements in a linked list at that index.
      3. Final Answer:

        Store multiple elements in a linked list at the same index when collisions occur -> Option C
      4. Quick Check:

        Chaining = linked list at index [OK]
      Hint: Chaining means linked lists at each index [OK]
      Common Mistakes:
      • Thinking chaining resizes the table
      • Confusing chaining with open addressing
      • Assuming collisions are ignored
      2. Which of the following is the correct way to declare a hash table with chaining in Go using slices of linked lists?
      easy
      A. var table [10]*Node
      B. var table = make([]*Node, 10)
      C. var table = make(map[int]*Node)
      D. var table = [10]Node{}

      Solution

      1. Step 1: Understand Go slice declaration for chaining

        Chaining uses a slice of pointers to linked list nodes for dynamic size.
      2. Step 2: Identify correct syntax

        Using make([]*Node, 10) creates a slice of 10 pointers to Node, suitable for chaining.
      3. Final Answer:

        var table = make([]*Node, 10) -> Option B
      4. Quick Check:

        Slice of pointers = make([]*Node, size) [OK]
      Hint: Use make([]*Node, size) for chaining table [OK]
      Common Mistakes:
      • Using arrays instead of slices limits flexibility
      • Using map instead of slice for chaining
      • Declaring slice without make causes nil slice
      3. Consider the following Go code snippet for inserting keys into a hash table with chaining. What will be the output after inserting keys 5 and 15 if the hash function is key % 10?
      type Node struct {
          key  int
          next *Node
      }
      
      func insert(table []*Node, key int) {
          index := key % 10
          newNode := &Node{key: key}
          newNode.next = table[index]
          table[index] = newNode
      }
      
      func printTable(table []*Node) {
          for i, node := range table {
              print(i, ": ")
              for n := node; n != nil; n = n.next {
                  print(n.key, " -> ")
              }
              println("nil")
          }
      }
      
      func main() {
          table := make([]*Node, 10)
          insert(table, 5)
          insert(table, 15)
          printTable(table)
      }
      medium
      A. 15 -> 5 -> nil at index 5
      B. 15 -> nil at index 5 and 5 -> nil at index 5
      C. 5 -> nil at index 5 and 15 -> nil at index 5
      D. 5 -> 15 -> nil at index 5

      Solution

      1. Step 1: Calculate hash index for keys 5 and 15

        Both 5 % 10 and 15 % 10 equal 5, so both keys go to index 5.
      2. Step 2: Trace insertion order in chaining

        First insert 5 at index 5, then insert 15 which becomes new head pointing to 5.
      3. Final Answer:

        15 -> 5 -> nil at index 5 -> Option A
      4. Quick Check:

        New nodes added at head = 15 first [OK]
      Hint: New nodes go to head of list in chaining [OK]
      Common Mistakes:
      • Assuming insertion appends at tail
      • Mixing up hash index calculation
      • Printing nodes in wrong order
      4. The following Go code tries to insert keys into a hash table with chaining but causes a runtime panic. What is the error?
      type Node struct {
          key  int
          next *Node
      }
      
      func insert(table []*Node, key int) {
          index := key % len(table)
          if table[index] == nil {
              table[index].key = key
          } else {
              newNode := &Node{key: key, next: table[index]}
              table[index] = newNode
          }
      }
      
      func main() {
          table := make([]*Node, 10)
          insert(table, 7)
      }
      medium
      A. Nil pointer dereference when assigning to table[index].key
      B. Index out of range error due to wrong hash calculation
      C. Infinite loop in insert function
      D. Compilation error due to missing return statement

      Solution

      1. Step 1: Analyze the nil check and assignment

        If table[index] is nil, trying to assign table[index].key causes nil pointer dereference.
      2. Step 2: Understand pointer initialization requirement

        Must create a new Node and assign pointer to table[index] before setting key.
      3. Final Answer:

        Nil pointer dereference when assigning to table[index].key -> Option A
      4. Quick Check:

        Assigning field on nil pointer causes panic [OK]
      Hint: Initialize pointer before assigning fields [OK]
      Common Mistakes:
      • Ignoring nil pointer before field assignment
      • Assuming slice elements auto-initialize
      • Confusing compile error with runtime panic
      5. You want to implement a hash table with chaining in Go that supports deletion of keys. Which approach correctly removes a key from the chain at the hashed index?
      hard
      A. Replace the key with zero but keep the node in the list
      B. Set the entire linked list at the index to nil to remove the key
      C. Resize the hash table and rehash all keys except the one to delete
      D. Traverse the linked list at the index, keep track of previous node, and adjust pointers to skip the node with the key

      Solution

      1. Step 1: Understand deletion in chaining

        To delete a key, find it in the linked list and remove its node by adjusting pointers.
      2. Step 2: Why other options fail

        Setting entire list to nil removes all keys, replacing key with zero keeps node, resizing is inefficient.
      3. Final Answer:

        Traverse the linked list at the index, keep track of previous node, and adjust pointers to skip the node with the key -> Option D
      4. Quick Check:

        Proper deletion adjusts pointers in linked list [OK]
      Hint: Delete by pointer adjustment in linked list [OK]
      Common Mistakes:
      • Removing entire chain instead of one key
      • Leaving empty nodes with zero keys
      • Rehashing unnecessarily on deletion
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