🚀 Mastering Array Rotation in Java: Algorithms and Time Complexities 🕰️
Published: 2023-10-03 14:53:50

Are you ready to take your Java programming skills to the next level? Let's dive into the fascinating world of array manipulation as we tackle the "Rotate an Array" problem with finesse! 🌟


![Image description](https://dev-to-uploads.s3.amazonaws.com/uploads/articles/rgdftkjjovm20ktoufn4.png)



**Problem Statement**
Imagine you have an array of integers, and you want to perform a cyclic right rotation on it by a certain number of positions, say k. Sounds tricky? Don't worry; we've got some nifty Java solutions for you!

**Example:**
Given the input array [1, 2, 3, 4, 5, 6, 7] and k = 3, the expected output is [5, 6, 7, 1, 2, 3, 4]. Magic, right? 🪄

**🌟 Solution 1: The Brute Force Approach 🌟**
Time Complexity: O(n * k)
In this approach, we roll up our sleeves and rotate the array one step at a time for k steps. It might not be the most efficient solution, but it's a great way to start understanding the problem.

```
// Java code for rotating an array using brute force
    public void rotate(int[] nums, int k) {
        int n = nums.length;
        k = k % n; // handle cases where k > n
        for (int i = 0; i < k; i++) {
            int temp = nums[n - 1];
            for (int j = n - 1; j > 0; j--) {
                nums[j] = nums[j - 1];
            }
            nums[0] = temp;
        }
    }
```


**🚀 Solution 2: Using Extra Space 🚀**
Time Complexity: O(n)

```
// Java code for rotating an array using extra space
public void rotate(int[] nums, int k) {
    int n = nums.length;
    int[] result = new int[n];
    for (int i = 0; i < n; i++) {
        result[(i + k) % n] = nums[i];
    }
    System.arraycopy(result, 0, nums, 0, n);
}

```
Next up, we take a more space-efficient approach. We create a new array and elegantly copy the rotated elements to it. This solution offers a significant improvement in time complexity.

**🌪️ Solution 3: Reversal Algorithm 🌪️**
Time Complexity: O(n)

In this solution, we take a smart approach. We reverse the array in parts: first, the entire array, then the first k elements, and finally the remaining elements. This results in an efficient solution with an optimal time complexity.

```
// Java code for rotating an array using the reversal algorithm
public void rotate(int[] nums, int k) {
    int n = nums.length;
    k = k % n;
    reverse(nums, 0, n - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, n - 1);
}

private void reverse(int[] nums, int start, int end) {
    while (start < end) {
        int temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start++;
        end--;
    }
}

```
**🎉 Conclusion 🎉**
You've just unlocked three powerful techniques to tackle the "Rotate an Array" problem in Java! Each solution comes with its own time complexity, allowing you to choose the best fit for your specific problem and performance requirements.

- Brute Force Approach (Time Complexity: O(n * k)): Simple and intuitive.
- Using Extra Space (Time Complexity: O(n)): Space-efficient and effective.
- Reversal Algorithm (Time Complexity: O(n)): Smart and optimal.

Feel free to ask questions, share your thoughts, or dive deeper into these solutions in the comments section below! 🤓

Happy coding, and remember, the sky's the limit when you master your algorithms! ✨