Insertion Sort MCQ Quiz in मराठी - Objective Question with Answer for Insertion Sort - मोफत PDF डाउनलोड करा

The correct answer is Unsupervised Learning.

Key Points

1. Supervised Learning:

• In supervised learning, the algorithm is trained on a labeled dataset, where the input data is paired with corresponding output labels.
• The goal is to learn a mapping function from input to output so that the algorithm can make predictions or classifications on new, unseen data.

2. Unsupervised Learning:

• In unsupervised learning, the algorithm is given data without explicit instructions on what to do with it.
• The algorithm tries to find patterns, relationships, or structures within the data on its own.
• Clustering algorithms, like k-Means, fall under unsupervised learning because they group similar data points together without using labeled output information.

3. Semi-supervised Learning:

• Semi-supervised learning is a combination of supervised and unsupervised learning.
• It involves a dataset that contains both labeled and unlabeled examples.
• The algorithm is trained on the labeled data, and then it tries to make predictions on the unlabeled data by leveraging the patterns learned from the labeled data.

4. Reinforcement Learning:

• Reinforcement learning involves an agent interacting with an environment and learning to make decisions by receiving feedback in the form of rewards or punishments.
• The agent learns to take actions that maximize the cumulative reward over time.
• Unlike supervised learning, where the algorithm is provided with explicit labeled examples, in reinforcement learning, the algorithm learns by trial and error.

In the case of the k-Means algorithm, it is unsupervised learning because it doesn't rely on labeled output data. Instead, it aims to partition the input data into clusters based on similarity, without using predefined class labels.

The correct answer is option 4.

Concept:

Insertion sort:

 Insertion sort is a basic sorting algorithm that operates in the same manner that you arrange cards in your hands. The array is divided into two halves, one sorted and one not. Values from the unsorted section are selected and placed in the sorted section at the appropriate location.

Explanation:

The given array is,

Input: 20, 16, 12, 8, 4, 1

Output:1, 4, 8, 12, 16, 20

Pass 1:
Consider the key is 16 and move the elements of array index 1 to 0
swaps are required because 20 is greater than 16.
Swap the elements 20 and 16
16, 20, 12, 8, 4, 1
No swaps are required because till array 1 index is sorted in order.
16, 20, 12, 8, 4, 1
Total swaps are=1
Pass 2:
Consider the key is 12 and move the elements of array index 2 to 0
swaps are required because 20 is greater than 12.
Swap the elements 20 and 12
16, 12, 20, 8, 4, 1
swaps are required because 16 is greater than 12.
Swap the elements 16 and 12
12, 16, 20, 8, 4, 1
No swaps are required because till the array 2 index is sorted in order.
12, 16, 20, 8, 4, 1
Total swaps are=3
Pass 3:
Consider the key is 8 and move the elements of array index 3 to 0
swaps are required because 20 is greater than 8.
Swap the elements 20 and 8
12, 16, 8, 20, 4, 1
swaps are required because 16 is greater than 8.
Swap the elements 16 and 8
12, 8, 16, 20, 4, 1
swaps are required because 12 is greater than 8.
Swap the elements 12 and 8
8, 12, 16, 20, 4, 1
No swaps are required because till array 3 index is sorted in order.
8, 12, 16, 20, 4, 1
Total swaps are=6
Pass 4:
Consider the key is 4 and move the elements of array index 4 to 0
swaps are required because 20 is greater than 4.
Swap the elements 20 and 4
8, 12, 16, 4, 20, 1
swaps are required because 16 is greater than 4.
Swap the elements 16 and 4
8, 12, 4, 16, 20, 1
swaps are required because 12 is greater than 4.
Swap the elements 12 and 4
8, 4, 12, 16, 20, 1
swaps are required because 8 is greater than 4.
Swap the elements 8 and 4
4, 8, 12, 16, 20, 1
No swaps are required because till array 4 index is sorted in order.
4, 8, 12, 16, 20, 1
Total swaps are=10
Pass 5:
Consider the key is 1 and move the elements of array index 5 to 0
swaps are required because 20 is greater than 1.
Swap the elements 20 and 1
4, 8, 12, 16, 1, 20
swaps are required because 16 is greater than 1.
Swap the elements 16 and 1
4, 8, 12, 1, 16, 20
swaps are required because 12 is greater than 1.
Swap the elements 12 and 1
4, 8, 1, 12, 16, 20
swaps are required because 8 is greater than 1.
Swap the elements 8 and 1
4, 1, 8, 12, 16, 20
swaps are required because 4 is greater than 1.
Swap the elements 4 and 1
1, 4, 8, 12, 16, 20
No swaps are required because till array 5 index is sorted in order.
1, 4, 8, 12, 16, 20
Total swaps are=15
The sorted array is
1, 4, 8, 12, 16, 20

Hence the correct answer is 4, 8, 12, 16, 20, 1.

The correct answer is option 3.

Concept:

The given example is a player playing a card game and sorting the cards. The Player first picks one card then picks the next card and put it after the first card if it is bigger or before the first card if it is smaller; then he picks another card and inserts it into its proper position.

A player follows the Insertion sort, It is a straightforward sorting algorithm that works similarly to how you arrange cards in your hands. The cards are divided into two parts: sorted and unsorted. The values of cards from the unsorted component are selected and placed in the sorted part in the proper order.

Example:

Consider the cards already sorted with 2, 4, 5, and 10. An unsorted card is 7 and is placed in the right position of a set of cards. Here card 7 is compared with 2, 4, 5, and 10 placed card 7 before 10. 

Hence the correct answer is Insertion Sort.

Additional Information

•  The selection sort algorithm sorts an array by continually picking the smallest member from the unsorted segment and placing it at the beginning (in ascending order). In a given array, the method keeps two subarrays. This is the subarray that has previously been sorted. The remaining unsorted subarray.
• Bubble Sort is the most basic sorting algorithm, which operates by exchanging neighbouring components in the wrong order repeatedly.

The correct answer is option 5.

Key Points

We have an array A with n distinct integers and each element is at most k away from its position in the sorted array. And we need to move the elements to their actual positions so we need to compare at most k elements are in the sorted part of the array. For total n elements, each element required k comparison so Total (NK) comparisons are required and n swappings are required. So worst-case number of comparisons is t(n) = Θ(NK).  It exactly works like insertion sort.

Hence the correct answer is t(n) = Θ(nk).