# 1. Introduction

The "Rotate Array" problem involves rotating the elements of an array to the right by k steps, where k is a non-negative integer. This problem tests the ability to manipulate array indices and understand modulo operations for circular rotations.

# 2. Problem

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.

# 3. Solution in C++

``````void rotate(vector<int>& nums, int k) {
k = k % nums.size();
reverse(nums.begin(), nums.end());
reverse(nums.begin(), nums.begin() + k);
reverse(nums.begin() + k, nums.end());
}
``````

### Explanation:

1. Calculate k modulo the size of the array to handle cases where k is larger than the array size.

2. Reverse the entire array.

3. Reverse the first k elements.

4. Reverse the remaining n-k elements.

5. This sequence of reversals achieves the desired right rotation.

# 4. Solution in Java

``````public void rotate(int[] nums, int k) {
k = k % nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 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--;
}
}
``````

### Explanation:

1. Compute k modulo the array's length.

2. Reverse the entire array.

3. Reverse the first k elements.

4. Reverse the rest of the array.

5. This approach effectively rotates the array to the right by k steps.

# 5. Solution in Python

``````def rotate(nums, k):
n = len(nums)
k = k % n
nums[:] = nums[n-k:] + nums[:n-k]
``````

### Explanation:

1. Calculate k modulo the length of nums to normalize k.

2. Slice the array into two parts: nums[n-k:] and nums[:n-k].

3. Concatenate these slices in reverse order to achieve the rotation.

4. This slicing and concatenating result in the array being rotated to the right by k steps.