Running time of “unique1” and “unique2” algorithms: As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook. Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them. The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates. So, it takes the worst-case running time of O ( n 2 ) to return the unique elements. The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other. It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O ( n log n ) time.

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Answer 1

Question

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

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Answer 2

Question

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

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Check out a sample textbook solution

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Answer 4

Question

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

Expert Solution & Answer

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Answer 5

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

Answer 6

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

Answer 7

Chapter 4, Problem 63P

Program Plan Intro

Running time of “unique1” and “unique2” algorithms:

As referred to the code fragment 4.7 shows the algorithm for “unique1” and code fragment 4.8 shows the algorithm for “unique2” in the textbook.

Both these algorithms solve the element uniqueness problem. But the running time of algorithm varies between them.

The algorithm for “unique1” solves the uniqueness by comparing all the distinct pairs. It uses two nested for loop to find out the duplicates.

So, it takes the worst-case running time of O(n2) to return the unique elements.

The algorithm of “unique2” solves the uniqueness by placing the duplicate element next to each other.

It includes a single pass on the array to sort the consecutive duplicates. It takes the worst-case running time of O(nlog n) time.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 4 - Prob. 1R Ch. 4 - The number of operations executed by algorithms A...Ch. 4 - The number of operations executed by algorithms A...Ch. 4 - Prob. 4R Ch. 4 - Prob. 5R Ch. 4 - Prob. 6R Ch. 4 - Prob. 7R Ch. 4 - Prob. 8R Ch. 4 - Prob. 9R Ch. 4 - Prob. 10R

Solution Summary: The author compares the running time of the algorithms "unique1" and "Unique2" in the book. Both solve the element uniqueness problem.

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