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Finding the Kth smallest element, the predecessor, may require traveling down the height of the tree, making the running time O(h).

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In-order is one type of DFS.

There are three types of depth-first traversal: pre-order, in-order, and post-order.

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The order: 76, 15, 88, 73, 17, 80, 61, 84 will result the given AVL

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→ Topological sort of DAG takes O(V+E) time.

→ Bellman ford single source shortest path takes O(VE) time

→ Floyd Warshall’s all pair shortest path algorithm takes O(V3) time.

→ Kruskal’s MST algorithm

• The edges are maintained as a min heap.
• The next edge can be obtained in O(logE) time if the graph has E edges.
• Reconstruction of heap takes O(E) time.
• So, Kruskal’s Algorithm takes O(ElogE) time.
• The value of E can be at most O(V^2).
• So, O(logV) and O(logE) are the same.

General Kruskal’s algorithm: Sorted edges will take O(ElogV) and unsorted edges will take O(ElogE)

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• BigOh(O) denotes upper bound or worst case
• BigOmega() denotes lower bound or best case
• BigTheta(𝛉) denotes Average bound or average case

Solution

The final sequence is a-b, b-c, c-i, c-f, f-g, g-h, c-d, d-e

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• A hash function H accepts a variable-length block of data M as input and produces a fixed-size hash value h = H(M).
• In general terms, the principal object of a hash function is data integrity. A change to any bit or bits in M results, with high probability, in a change to the hash code.
• Cryptographic hash function should have following two properties:

• The one-way property: It should be computationally infeasible to find the data object that maps to a pre-specified hash result.
• Collision-free property: No two data objects should map to the same hash result

Because of these characteristics, hash functions are often used to determine whether or not data has changed.

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I. TRUE: Undirected graphs do not have cross edges in DFS. But can have cross edges in directed graphs.

II. FALSE: Just draw a triangle ABC. Source is A. Vertex B and C are at the same level at distance 1. There is an edge between B and C too. So here |i - j| = |1 - 1| = 0.

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P: Binary search tree → divide and conquer technique.

Q: 0/1 knapsack problem → Dynamic programming approach.

R: Huffman Coding → Greedy

S: Depth-first Search → Stack

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Perimeter = 36 cm

AB + BC + AC = 36 -------(i)

⟶ From the question, we know

BC = 2AB - 4 -------(ii)

AC = AB + 8 -------(iii)

⟶ Put (ii) and (iii) in (i), we get

(AB) + (2AB - 4) + (AB + 8) = 36

4AB = 32

AB = 8

BC = 12

AC = 16

∴ Smallest side is AB = 8 cm

Solution

There are four types of tenses that together make up Present Tense, namely:

The Simple Present Tense is used:To express habits, universal truths, repeated actions or fixed events.For example :The school bus picks up the students at 6 am.

The Present Perfect Tense is used to indicate a link between the present and the past. It is used to describe:An action that started in the past and continues in the present. For example: He has worked in this company since 2004.

The Present Continuous Tense is made from the present form of the verb ‘be’ and the -ing form of a verb. It is used:To describe an action that is happening at the moment of speaking. For example:The children are playing in the field.

The Present Perfect Continuous Tense is used for an action that started in the past and has continued up to the present moment. It is used:To describe an action that started in the past and has continued up to the present. For example: She has been singing for over two hours.

Solution

Except option b) when we use the other options in the sentence, they are grammatically correct.

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Inchoate means only partly in existence; imperfectly formed.

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Crowdfunding is so effective because campaigns reach their goals by receiving many small donations from a large group of supporters, rather than soliciting gifts from a few major contributors.

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Multi-valued attribute: double oval

Derived attribute: dashed oval

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Wait-Die Scheme: In this scheme, If a transaction requests for a resource that is locked by another transaction, then the DBMS simply checks the timestamp of both transactions and allows the older transaction to wait until the resource is available for execution.

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Suppose timestamp ordering protocol is used to control concurrent access. In this context a transaction Ti issues a read(Q) then, if Timestamp(Ti) ≤ Write-timestamp(Q), the read operation is rejected, and Ti is rolled back.

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Simple random sampling provides each member of the population with an equal chance of selection, ensuring fairness in sample representation.

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Each transaction is associated with 10 products, meaning each transaction generates 10 rows in the fact table. Therefore, the total number of rows is:

1,000,000 * 10 = 10,000,000 rows

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In a star schema, joining the fact table with a dimension table does not increase the number of rows. So, the query will return:

500,000 rows

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When the data has noise and overlapping points, there is a problem in drawing a clear hyperplane without misclassifying.

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Var(2X-3Y) = 34

= 4Var(X)+9Var(Y)-12Cov(X, Y)

= 4(1)+9(4)-12Cov(X, Y) = 34

∴ Cov(X, Y)=1/2

Solution

In binary classification, there are typically two classes or categories to predict, such as yes/no or spam/ham. In a neural network for binary classification, a single output node is often used, where the output represents the probability or confidence score of belonging to one of the two classes

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The IQR is essentially the range of the middle 50% of the data. Since it uses the middle 50%, therefore it is not affected by the outliers.

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The F1-score is a common evaluation metric for classification models, particularly when dealing with imbalanced datasets. It combines precision and recall, making it a robust choice when there is an unequal distribution of classes. It provides a balanced measure of a model's accuracy in such cases.

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Overfitting happens when the model is too complex relative to the data. Adding more layers (D) generally increases the model's capacity to memorize the training data, thereby exacerbating overfitting rather than reducing it.

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K-1=5-1=4

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UCS prioritizes nodes with the lowest cumulative path cost, expanding them before others with higher costs.

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BFS is optimal for finding paths with the fewest edges but does not account for path cost, which is essential in graphs with varying edge weights.

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Marginal Inference: This involves calculating the probability distribution of a subset of variables (variables of interest) in a Bayesian network, without requiring specific values for other variables. It computes the marginal probability by summing or integrating over all possible values of the other variables.

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Uncertainty in decision-making refers to the lack of complete knowledge or clarity about future events or outcomes. It reflects ambiguity in predicting how situations will unfold, which makes it challenging to make informed decisions.

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LetX be the variable force values and Y be the dependent variable extension values. The coefficient of correlation is given by:

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Hence the Chi-square value χ^2 = 4.36

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et the speed of cars is represented by a random variable ‘x’.

Now, given mean, μ = 90 and standard deviation, σ = 10.

To find: Probability that x is higher than 100 or P(x > 100)

By using the transformation equation, we know;

z = (X – μ) / σ

Hence,

For x = 100 , z = (100 – 90) / 10 = 1

P(x > 90) = P (z > 1) = [total area] – [area to the left of z = 1]

P(z > 1) = 1 – 0.8413 = 0.1587

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|A| = 3,

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Rank(A) ≤ min(4,5), Rank ≤ 4

But System is inconsistent ,so rank = 4 is not possible.

Therefore highest possible rank of A is 3.

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Only one matrix has an Eigen value of 1

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by Gauss elimination Process

Now, for infinite solution last row must be completely zero

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y = | 2 - 3x |

Since 2-3x and 3x-2 are polynomials, these are continuous at all points. The only concern is

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The function adds n and recursively calls itself with n // 2 until n becomes 0.

The sequence is 10 + 5 + 2 + 1 = 15

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Mutating a dictionary (x[key + 1]) during iteration raises a RuntimeError.

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The code correctly removes duplicates while keeping the order intact by checking if an item is not in the unique_items list before appending it.

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The GATE only visits nodes connected to the starting node. Nodes 4 and 5 are in a separate disconnected component.

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While generators are often more memory-efficient than iterators, this isn't always the case. It depends on the specific implementation and use case. In many scenarios, iterators can be equally efficient.