Searching MCQ Quiz in मल्याळम - Objective Question with Answer for Searching - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Homogeneous data structure (HDS):

• HDS that contains only similar type of data like only integers or only float values.
• Basic example of Homogeneous data structure is Array.

Explanation

In question we required ordered finite HDS which is nothing but Array but Array is not present in the

Options so we have to choose best option among them Let’s take option wise.

Option 1:  Linked list

Linked list is HDS but not finite because linked list size is decided on Run time.

Option 2: Graph

Graph is HDS but not finite because Graph size is decided on Run time.

Option 3:  Tree

Tree is HDS but not finite because Tree size is decided on Run time.

Option 4:  Hash Table

Hash table is ordered and finite data structure (Size is decided on Compile time ).

Hence option 4 is the Best option to choose for the answer.

Binary search algorithm:

Binary search algorithm is used to find an element in an already sorted array.

STEPS 1:

It finds the middle element of the array and compare the element with the element to be searched, if

it matches then return true.

STEPS 2:

If not, then divide the array into two halves in which if element to be search is less than middle

element then search takes place in left part otherwise search in right half.

STEP 3:

Repeat this process, until you get the element.

Explanation:

Best case of Binary search occurs when element to be searched is the middle element of the array.

Then in that case, it just simply compares the element with the middle element, if matches then return

true.  Comparing with the one element will take only O(1) time complexity.

Best Case: Search 35 in below give array

T(n) = O(1)

The correct answer is option 1.

Concept:

Linear search:

A linear search, often known as a sequential search, is a technique for locating an element in a list. It systematically verifies each element of the list until a match is discovered or the entire list has been searched.

Algorithm:

linear_search(int a[], int n, int X)

{ 
  for (int i = 0; i < n; i++)  
    {  
        if (a[i] == X)  
        return i+1;  
    } 

}

Explanation:

Best case:

The best case in the Linear Search Algorithm happens when the item to be found is at the beginning of the Array. If the element is found at starting of the array the linear search successful and comes out from the loop. Hence this behavior takes less time when compared to other behaviors.

Worst-case:

In the Linear Search Algorithm, the worst-case scenario happens when the item to be found is at the end of the Array or the element is not present in the array. Hence the linear search compares each element till the end. So it takes maximum time behavior in linear search.

Average case:

In the Linear Search Algorithm, the average scenario happens when the item to be found is somewhere in the center of the Array.

Hence the correct answer is a searching element is at beginning of the array.

Key Points

 The average number of comparisons in a sequential search is \({(N+1)} \over 2\) where N is the size of the array.

If the element is in the 1st position, the number of comparisons will be 1,

the element is in the 2^nd position, the number of comparisons will be 2,

the element is in the 3^rd position, the number of comparisons will be 3,

the element is in the 4^th position, the number of comparisons will be 4, .....

if the element is in the last position, the number of comparisons will be N.

So Total number of comparisons are= (1+2+3+4+5+...+N)

The total number of comparisons are=\({N(N+1)} \over 2\)

The average number of comparisons in a sequential search is \({N(N+1)} \over 2 \times N\)

Hence the correct answer is \({(N+1)} \over 2\)

The correct answer is Linear probing

Key Points

• Linear Probing:
  • When a collision occurs (two elements hash to the same location), linear probing involves placing the collided element in the next available (unoccupied) slot in the hash table.
  • If that slot is also occupied, it continues to search for the next empty slot in a linear fashion (moving one slot at a time) until an empty slot is found.
  • Linear probing can lead to clustering, where consecutive elements form clusters in the hash table.

Additional Information

• Quadratic Probing:
  • Similar to linear probing, but instead of moving one slot at a time, quadratic probing uses a quadratic function to determine the next position to check.
  • If the slot at the hash index is occupied, it checks slots at positions incremented by successive squares.
  • Quadratic probing helps to reduce clustering compared to linear probing.
• Separate Chaining:
  • Instead of placing collided elements in the next available slot in the hash table, separate chaining involves maintaining a linked list at each slot in the hash table.
  • When a collision occurs, the collided elements are added to the linked list at that slot.
  • Each slot contains a separate data structure (like a linked list or a tree) to handle collisions.
• Double Hashing:
  • In double hashing, a secondary hash function is used to determine the step size between probe attempts.
  • If a collision occurs, the algorithm uses the secondary hash function to calculate a new index to probe.
  • This helps to avoid clustering and provides a more systematic way to find an empty slot.

The correct option is Search by Hashing.

CONCEPT:

In linear search, every element of a given list is compared one by one with the given key without skipping any element.

It is useful when we need to search for an item in a small unsorted list, but the time taken to search the list increases as the size of the list increases.

For example, consider a list of length 5 and the key element is present at the end of this list. Number of comparisons required to search the key element = size of list i.e 5

If we increase the size of the same list (say 15) and the key element is present at the end of this list. Number of comparisons required to search the key element = size of list i.e 15

In binary search, the key to be searched is compared with the middle element of a sorted list, If the element at the middle position:

i)  Matches the key the search is successful

ii) Is greater than the key then the key element may be present in the left part of the list

iii) Is smaller than the key, then the key element may be present in the right part of the list

This process continues till the element is found or the list is traversed completely.

Thus the time taken to search the list increases as the size of the list increases. But it will not be as large as the time required by linear search

Hash-based searching requires only one key comparison to find the position of a key, provided every element is present at its designated position decided by a hash function. 

For example, consider a list of length 5 and the hash function: h(element) = element % size(list)

A hashing function is a mathematical function that generates unique results for each unique value supplied to the hash function in constant time.

For example: Consider a list of length 5 and if we want to search for key = 12, the index returned by the hash function is h(12) = 12 % 5  = 2  and requires only one key comparison to find the key at that index

Similarly increasing the size of the list (say 15) and if we want to search for key = 12, the index returned by the hash function is h(12) = 12 % 5  = 12  and requires only one key comparison to find the key at that index

Thus it is independent of the length of the list.

Formula:

Expected number of probes in unsuccessful search = \(\frac{1}{{1 - p}}\)

Expected number of probes in successful search = \(\frac{1}{p}\ln \left( {\frac{1}{{1 - p}}} \right)\)

\(p = \frac{n}{m}\)

where p is load factor.

Calculation:

\(\frac{1}{{1 - p}} = 5\)

1 = 5 – 5p 

5p = 4

\(p = \frac{4}{5}\)

\(\frac{1}{p}\ln \left( {\frac{1}{{1 - p}}} \right) = \frac{5}{4}\ln \left( {\frac{1}{{1 - \frac{4}{5}\;}}} \right) = \frac{5}{4}\ln \left( 5 \right) = \frac{5}{4} \times 1.609 = 2.011\)

So, number of probes in successful search is 2.01

Answer: 1.5

Explanation:

In a hash-table with chaining and uniform hashing, the expected number of comparisons in an unsuccessful search is given by

α, where α is the load factor

\(\alpha = \frac{m}{n} = \frac{{60}}{{40}} = 1.5\)

Here, So, expected

number of comparisons for an unsuccessful search = 1.5

Worst case of the given problem is when a single 0 is followed by all 1’s i.e.

0111111111111……

Also, worst case can also be considered as when all 0’s are followed by single 1.

000000000………………1

Since, in both the cases there is a sorted sequence of 0 and 1 and for sorted sequence binary search is preferred.

At each stage, \(\frac{{low + high}}{2}\) element index is compared and if it is 1, search towards left and if it is 0 search towards right.