Which are the correct low and high partitions of the array {10, 20, 80, 40, 90, 70} assuming the middle element as pivot? O {10,20] and [80,40,90,70} {10,20,40) and {80,70,90}

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
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O(n3) and  {10,20,40) and {80,70,90} are incorrect

### Question:
Which are the correct low and high partitions of the array {10, 20, 80, 40, 90, 70} assuming the middle element as pivot?

### Options:
1. ⭘ {10,20} and {80,40,90,70}
2. ⭘ {10,20,70,40} and {90,80}
3. ⭘ {10,40,20} and {80,90,70}
4. ⭙ {10,20,40} and {80,70,90}

### Answer:
The correct option is:
- {10,20,40} and {80,70,90}

### Explanation:
When partitioning an array using the middle element as a pivot, the elements less than or equal to the pivot are grouped into the "low" partition, and the elements greater than the pivot are grouped into the "high" partition. Here, the middle element in the array {10, 20, 80, 40, 90, 70} is 40.

So, the array is partitioned as follows:
- Low partition: {10, 20, 40}
- High partition: {80, 70, 90}

The correct answer is {10,20,40} and {80,70,90}.
Transcribed Image Text:### Question: Which are the correct low and high partitions of the array {10, 20, 80, 40, 90, 70} assuming the middle element as pivot? ### Options: 1. ⭘ {10,20} and {80,40,90,70} 2. ⭘ {10,20,70,40} and {90,80} 3. ⭘ {10,40,20} and {80,90,70} 4. ⭙ {10,20,40} and {80,70,90} ### Answer: The correct option is: - {10,20,40} and {80,70,90} ### Explanation: When partitioning an array using the middle element as a pivot, the elements less than or equal to the pivot are grouped into the "low" partition, and the elements greater than the pivot are grouped into the "high" partition. Here, the middle element in the array {10, 20, 80, 40, 90, 70} is 40. So, the array is partitioned as follows: - Low partition: {10, 20, 40} - High partition: {80, 70, 90} The correct answer is {10,20,40} and {80,70,90}.
**Question**: What is the worst-case time complexity for the following code fragment?

```cpp
int x = 0;
if (n % 2 == 0) {
    for (int i = 0; i < n * n * n * n; i++) {
        x++;
    }
} else {
    for (int i = 0; i < n * n * n; i++) {
        x++;
    }
}
```

**Options**:
- O(n²)
- O(n³) (selected)
- O(n)
- O(n⁷)
- O(n⁴)

**Explanation**:
The code includes an `if` statement that evaluates whether `n` is even or odd. If `n` is even (`n % 2 == 0`), the code executes a `for` loop that iterates `n * n * n * n` times. If `n` is odd, it executes a `for` loop that iterates `n * n * n` times.

### Detailed Analysis:
- **If `n` is even**: 
    The loop runs from `0` to `n^4`, resulting in `n^4` iterations.
    \[
    T(n) = O(n^4)
    \]

- **If `n` is odd**: 
    The loop runs from `0` to `n^3`, resulting in `n^3` iterations.
    \[
    T(n) = O(n^3)
    \]

Thus, the worst-case time complexity is \(O(n^4)\) since \(O(n^4)\) is larger than \(O(n^3)\).

However, based on the selected option in the image, the correct answer appears to be marked as \(O(n^3)\), which suggests a possible misinterpretation. The correct answer should be \(O(n^4)\), taking into account the `if` branch.

### Graphical Representation:
Unfortunately, no graphical data (like charts or diagrams) are present here to further illustrate the code’s time complexity.

Updated Correct Answer:
- O(n⁴)
Transcribed Image Text:**Question**: What is the worst-case time complexity for the following code fragment? ```cpp int x = 0; if (n % 2 == 0) { for (int i = 0; i < n * n * n * n; i++) { x++; } } else { for (int i = 0; i < n * n * n; i++) { x++; } } ``` **Options**: - O(n²) - O(n³) (selected) - O(n) - O(n⁷) - O(n⁴) **Explanation**: The code includes an `if` statement that evaluates whether `n` is even or odd. If `n` is even (`n % 2 == 0`), the code executes a `for` loop that iterates `n * n * n * n` times. If `n` is odd, it executes a `for` loop that iterates `n * n * n` times. ### Detailed Analysis: - **If `n` is even**: The loop runs from `0` to `n^4`, resulting in `n^4` iterations. \[ T(n) = O(n^4) \] - **If `n` is odd**: The loop runs from `0` to `n^3`, resulting in `n^3` iterations. \[ T(n) = O(n^3) \] Thus, the worst-case time complexity is \(O(n^4)\) since \(O(n^4)\) is larger than \(O(n^3)\). However, based on the selected option in the image, the correct answer appears to be marked as \(O(n^3)\), which suggests a possible misinterpretation. The correct answer should be \(O(n^4)\), taking into account the `if` branch. ### Graphical Representation: Unfortunately, no graphical data (like charts or diagrams) are present here to further illustrate the code’s time complexity. Updated Correct Answer: - O(n⁴)
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