public void mystery(int[] arr) {
int n = arr.length;
if (n == 0) {
System.out.println("Done.");
} else if (n % 2 == 0) {
int i = 0;
while (i < n) {
System.out.println(arr[i]);
i++;
}
} else {
for (int j = 0; j < n; j += 2) {
System.out.println("j is " + j);
System.out.println("sum is");
for (int k = 0; k < n - 1; k += 2) {
System.out.println(arr[j] + arr[k]);
}
}
}
}

Write the runtime of the worst case and give any input arr that will achieve the worst case.

### Analyzing the Code and Determining the Worst-Case Runtime

Let's break down the code to analyze the worst-case runtime.

```java
public void mystery(int[] arr) {
    int n = arr.length;
    if (n == 0) {
        System.out.println("Done.");
    } else if (n % 2 == 0) {
        int i = 0;
        while (i < n) {
            System.out.println(arr[i]);
            i++;
        }
    } else {
        for (int j = 0; j < n; j += 2) {
            System.out.println("j is " + j);
            System.out.println("sum is");
            for (int k = 0; k < n - 1; k += 2) {
                System.out.println(arr[j] + arr[k]);
            }
        }
    }
}
```

#### Step-by-Step Analysis:

1. **Base Case (Empty Array):**
   - If `n == 0`, it just prints `"Done."` and terminates. This has a constant runtime of \(O(1)\).

2. **Case 1: `n % 2 == 0` (Even Length Array):**
   - The program enters the `else if (n % 2 == 0)` block.
   - Inside this block, there is a `while` loop that iterates from `i = 0` to `i = n - 1`.
   - Each iteration involves printing an element of the array.
   - This loop runs \(n\) times, so the time complexity of this block is \(O(n)\).

3. **Case 2: `n % 2 != 0` (Odd Length Array):**
   - The program enters the `else` block.
   - There is a `for` loop with index `j` that starts from 0 and increments by 2, running until `j < n`.
   - This loop will run approximately \( \lceil n/2 \rceil \) times (since it iterates over every second element).
   - Inside this loop, there's another `for` loop with index `k`, which also increments by 2 and runs until `k < n-1`.
   - The inner loop will also run approximately \( \lceil (n-1)/2 \rceil \) times for each `j`.
   - The operations inside the nested loops involve printing values, which are \(O(1)\) operations.

   Therefore, the time complexity of this block is \(O\left(\left(\frac{n}{2}\right) \times \left(\frac{n-1}{2}\right)\right)\), which simplifies to \(O(n^2)\).

### **Worst-Case Runtime:**
The worst-case runtime occurs when `n` is odd because the time complexity of the odd case is \(O(n^2)\), which is greater than the \(O(n)\) complexity when `n` is even.

### **Example of Worst-Case Input:**
To achieve the worst-case scenario, provide an array with an odd number of elements. For example:

```java
int[] arr = {1, 2, 3, 4, 5};
```

This input will cause the function to run in \(O(n^2)\) time.

### Summary:
- **Worst-case runtime:** \(O(n^2)\)
- **Worst-case input example:** Any array with an odd number of elements, such as `{1, 2, 3, 4, 5}`.

Would you like more details on any part of this analysis or have any questions?

#### Follow-up Questions:
1. What would be the best-case scenario, and what is its runtime?
2. How does the complexity change if we modify the loop increments?
3. Can you rewrite the code to reduce the time complexity for odd-length arrays?
4. What is the space complexity of this function?
5. How would the runtime change if we used different data structures (e.g., `ArrayList` instead of an array)?

**Tip:** When analyzing nested loops, always consider the loop increments and how many times each loop will run to accurately determine the time complexity.

Explore the time complexity analysis of Java code involving nested loops. Understand how different loop structures affect performance, with a focus on scenarios where time complexity can reach O(n^2). Suitable for advanced programmers and computer science students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation