Space Complexity MCQ Quiz - Objective Question with Answer for Space Complexity - Download Free PDF

The correct answer is option 4.

Explanation: 

Here are some common Big O notations from the smallest to the largest complexity:

• O(1): Constant-time complexity. The running time is independent of the input size, it always takes the same time to compute.
• O(log n): Logarithmic time complexity. The running time grows logarithmically in proportion to the input size.
• O(n): Linear time complexity. The running time grows linearly with the size of the input.
• O(n log n): Log-linear or Linearithmic time complexity. Running time grows linearly and in the order of the logarithm of the size of the input.
• O(n^2): Quadratic time complexity. The running time is proportional to the square of the size of the input data.
• O(n^3): Cubic time complexity. The running time is proportional to the cube of the size of the input data.
• O(2^n): Exponential time complexity. The resources needed for algorithm execution double with each addition to the input data set.
• O(n!): Factorial time complexity. The running time grows in proportion to the factorial of the input size. This is typical of algorithms that solve problems by generating all possible combinations.

So when you say "all types of asymptotic complexity increasing order", we could be talking about a set of functions like:

• f1(n) = 1 (O(1))
• f2(n) = log n (O(log n))
• f3(n) = n (O(n))
• f4(n) = n log n (O(n log n))
• f5(n) = n^2 (O(n^2))
• f6(n) = n^3 (O(n^3))
• f7(n) = 2^n (O(2^n))
• f8(n) = n! (O(n!))

So, according to above explanation:

•  f1(n) = 2ⁿ
•  f2(n) = n^3/2
•  f3(n) = n log n 
•  f4(n) = n^{log n}

The functions in ascending order: n log n  <  n^3/2 < n^{log n}   <   2ⁿ

Answer: Option 3

Explanation: 

A measure of the amount of memory needed for an algorithm to execute is called Space efficiency or space complexity.

Additional Information

Time complexity: 

A measure of the amount of time needed for an algorithm to execute is called Time efficiency or Time complexity.

The correct answer is space efficiency.

Key Points

• Space efficiency is a measure of the amount of memory required for an algorithm to execute.
  • It indicates how the memory usage of an algorithm scales with the input size.
  • An algorithm with good space efficiency uses less memory, which is crucial for handling large data sets and optimizing performance.
  • Space efficiency is particularly important in systems with limited memory resources.
  • It helps in determining whether an algorithm can run within the available memory constraints.
  • Analyzing the space complexity of an algorithm involves understanding its memory consumption patterns and identifying potential areas for optimization.

Additional Information

• Space efficiency complements time efficiency, which measures the runtime performance of an algorithm.
• Both space and time efficiency are critical factors in the design and analysis of efficient algorithms.
• Commonly used notations to represent space complexity include Big O (O), Big Omega (Ω), and Big Theta (Θ).
• Examples of space-efficient algorithms include those that use in-place sorting or data structures that minimize memory overhead.

Answer: Option 3

Explanation: 

A measure of the amount of memory needed for an algorithm to execute is called Space efficiency or space complexity.

Additional Information

Time complexity: 

A measure of the amount of time needed for an algorithm to execute is called Time efficiency or Time complexity.

The correct answer is option 1.

Explanation: 

Here are some common Big O notations from the smallest to the largest complexity:

• O(1): Constant-time complexity. The running time is independent of the input size, it always takes the same time to compute.
• O(log n): Logarithmic time complexity. The running time grows logarithmically in proportion to the input size.
• O(n): Linear time complexity. The running time grows linearly with the size of the input.
• O(n log n): Log-linear or Linearithmic time complexity. Running time grows linearly and in the order of the logarithm of the size of the input.
• O(n^2): Quadratic time complexity. The running time is proportional to the square of the size of the input data.
• O(n^3): Cubic time complexity. The running time is proportional to the cube of the size of the input data.
• O(2^n): Exponential time complexity. The resources needed for algorithm execution double with each addition to the input data set.
• O(n!): Factorial time complexity. The running time grows in proportion to the factorial of the input size. This is typical of algorithms that solve problems by generating all possible combinations.

So when you say "all types of asymptotic complexity increasing order", we could be talking about a set of functions like:

• f1(n) = 1 (O(1))
• f2(n) = log n (O(log n))
• f3(n) = n (O(n))
• f4(n) = n log n (O(n log n))
• f5(n) = n^2 (O(n^2))
• f6(n) = n^3 (O(n^3))
• f7(n) = 2^n (O(2^n))
• f8(n) = n! (O(n!))

So, according to above explanation:

•  f1(n) = 2ⁿ
•  f2(n) = n^3/2
•  f3(n) = n log n 
•  f4(n) = n^{log n}

The functions in ascending order: n log n  <  n^3/2 < n^{log n}   <   2ⁿ

The correct answer is option 1.

Explanation: 

Here are some common Big O notations from the smallest to the largest complexity:

• O(1): Constant-time complexity. The running time is independent of the input size, it always takes the same time to compute.
• O(log n): Logarithmic time complexity. The running time grows logarithmically in proportion to the input size.
• O(n): Linear time complexity. The running time grows linearly with the size of the input.
• O(n log n): Log-linear or Linearithmic time complexity. Running time grows linearly and in the order of the logarithm of the size of the input.
• O(n^2): Quadratic time complexity. The running time is proportional to the square of the size of the input data.
• O(n^3): Cubic time complexity. The running time is proportional to the cube of the size of the input data.
• O(2^n): Exponential time complexity. The resources needed for algorithm execution double with each addition to the input data set.
• O(n!): Factorial time complexity. The running time grows in proportion to the factorial of the input size. This is typical of algorithms that solve problems by generating all possible combinations.

So when you say "all types of asymptotic complexity increasing order", we could be talking about a set of functions like:

• f1(n) = 1 (O(1))
• f2(n) = log n (O(log n))
• f3(n) = n (O(n))
• f4(n) = n log n (O(n log n))
• f5(n) = n^2 (O(n^2))
• f6(n) = n^3 (O(n^3))
• f7(n) = 2^n (O(2^n))
• f8(n) = n! (O(n!))

So, according to above explanation:

•  f1(n) = 2ⁿ
•  f2(n) = n^3/2
•  f3(n) = n log n 
•  f4(n) = n^{log n}

The functions in ascending order: n log n  <  n^3/2 < n^{log n}   <   2ⁿ

Answer: Option 3

Explanation: 

A measure of the amount of memory needed for an algorithm to execute is called Space efficiency or space complexity.

Additional Information

Time complexity: 

A measure of the amount of time needed for an algorithm to execute is called Time efficiency or Time complexity.

Answer: Option 3

Explanation: 

A measure of the amount of memory needed for an algorithm to execute is called Space efficiency or space complexity.

Additional Information

Time complexity: 

A measure of the amount of time needed for an algorithm to execute is called Time efficiency or Time complexity.

From Diagram: NP – complete ⋂ P = ϕ

NP-complete problems:

They are those for which no polynomial-time algorithm exists. We can say a problem is NP-complete if it is NP and belongs to NP-hard.

NP problems:

A problem is a member of NP class if there exists a non-deterministic machine which can solve it in polynomial time.

NP Hard:

A problem is called as NP hard if all the NP class problems are polynomial time reducible to that and as hard as any problem of NP class.

The correct answer is Option 4.

Concept:

We are given the following information:

• P3 is known to be NP-hard.
• P1 is known to be in the class of NP.

From this information, we can make the following conclusions:

• P3 is NP-hard: This means that P3 is at least as hard as the hardest problems in the class NP. Any problem in NP can be reduced to P3 in polynomial time. In other words, if we have an algorithm that solves P3 in polynomial time, we can use it to solve any problem in NP in polynomial time.
• P1 is in NP: This means that there exists a non-deterministic Turing machine that can solve problem P1 in polynomial time. NP is the class of decision problems that can be verified in polynomial time.

Explanation:

Now, let's analyze the answer options:

P = NP:

• This statement claims that all problems in NP can be solved in polynomial time.
• This statement has not been proven or disproven.
• It is one of the most significant open questions in computer science and remains unresolved.
• Therefore, we cannot conclude that P = NP based on the given information.

A ∈ P:

• This statement suggests that P1 (represented by A) is in the class P, which consists of problems solvable in polynomial time.
• While we know that P1 is in NP, we cannot directly conclude that it is in P.
• NP includes problems that may not be solvable in polynomial time.

P1 is NP-complete:

• This statement claims that P1 is both NP-hard and in NP.
• However, we do not have enough information to determine whether P1 is NP-complete.
• Just knowing that P3 is NP-hard and P1 is in NP does not guarantee that P1 is NP-complete.
• NP-completeness is a stronger condition that requires a reduction from all problems in NP to P1, not just P3.

P1 is NP-hard but not NP-complete:

• This statement correctly captures the information we have.
• Since P3 is NP-hard, any problem in NP can be reduced to P3.
• However, we do not have enough information to determine if P1 is NP-complete.

Hence, the correct answer is option 4.

The correct answer is option 1.

Explanation: 

Here are some common Big O notations from the smallest to the largest complexity:

• O(1): Constant-time complexity. The running time is independent of the input size, it always takes the same time to compute.
• O(log n): Logarithmic time complexity. The running time grows logarithmically in proportion to the input size.
• O(n): Linear time complexity. The running time grows linearly with the size of the input.
• O(n log n): Log-linear or Linearithmic time complexity. Running time grows linearly and in the order of the logarithm of the size of the input.
• O(n^2): Quadratic time complexity. The running time is proportional to the square of the size of the input data.
• O(n^3): Cubic time complexity. The running time is proportional to the cube of the size of the input data.
• O(2^n): Exponential time complexity. The resources needed for algorithm execution double with each addition to the input data set.
• O(n!): Factorial time complexity. The running time grows in proportion to the factorial of the input size. This is typical of algorithms that solve problems by generating all possible combinations.

So when you say "all types of asymptotic complexity increasing order", we could be talking about a set of functions like:

• f1(n) = 1 (O(1))
• f2(n) = log n (O(log n))
• f3(n) = n (O(n))
• f4(n) = n log n (O(n log n))
• f5(n) = n^2 (O(n^2))
• f6(n) = n^3 (O(n^3))
• f7(n) = 2^n (O(2^n))
• f8(n) = n! (O(n!))

So, according to above explanation:

•  f1(n) = 2ⁿ
•  f2(n) = n^3/2
•  f3(n) = n log n 
•  f4(n) = n^{log n}

The functions in ascending order: n log n  <  n^3/2 < n^{log n}   <   2ⁿ

Answer: Option 3

Explanation: 

A measure of the amount of memory needed for an algorithm to execute is called Space efficiency or space complexity.

Additional Information

Time complexity: 

A measure of the amount of time needed for an algorithm to execute is called Time efficiency or Time complexity.

Answer: Option 3

Explanation: 

A measure of the amount of memory needed for an algorithm to execute is called Space efficiency or space complexity.

Additional Information

Time complexity: 

A measure of the amount of time needed for an algorithm to execute is called Time efficiency or Time complexity.

The correct answer is O(1), O(1)

Key Points

• Inserting an element into an array at the end is an O(1) operation, meaning it takes constant time regardless of the size of the array. In other words, it doesn't matter if you're inserting one element or 100 elements, each insertion is a constant time operation, hence O(1).
• If you would have inserted at a specific position (not end) or removed from a specific position then it would be O(n) as it does involve shifting of elements.

The correct answer is option 4.

Explanation: 

Here are some common Big O notations from the smallest to the largest complexity:

• O(1): Constant-time complexity. The running time is independent of the input size, it always takes the same time to compute.
• O(log n): Logarithmic time complexity. The running time grows logarithmically in proportion to the input size.
• O(n): Linear time complexity. The running time grows linearly with the size of the input.
• O(n log n): Log-linear or Linearithmic time complexity. Running time grows linearly and in the order of the logarithm of the size of the input.
• O(n^2): Quadratic time complexity. The running time is proportional to the square of the size of the input data.
• O(n^3): Cubic time complexity. The running time is proportional to the cube of the size of the input data.
• O(2^n): Exponential time complexity. The resources needed for algorithm execution double with each addition to the input data set.
• O(n!): Factorial time complexity. The running time grows in proportion to the factorial of the input size. This is typical of algorithms that solve problems by generating all possible combinations.

So when you say "all types of asymptotic complexity increasing order", we could be talking about a set of functions like:

• f1(n) = 1 (O(1))
• f2(n) = log n (O(log n))
• f3(n) = n (O(n))
• f4(n) = n log n (O(n log n))
• f5(n) = n^2 (O(n^2))
• f6(n) = n^3 (O(n^3))
• f7(n) = 2^n (O(2^n))
• f8(n) = n! (O(n!))

So, according to above explanation:

•  f1(n) = 2ⁿ
•  f2(n) = n^3/2
•  f3(n) = n log n 
•  f4(n) = n^{log n}

The functions in ascending order: n log n  <  n^3/2 < n^{log n}   <   2ⁿ