**Approach**

We will use a three-step approach to solve this problem:

- Calculate the effective rotation index k: Since k can be greater than the length of the array, we need to find the effective rotation index k. We can do this by taking the modulus of k with the length of the array.
- Reverse the entire array: Once we have the effective rotation index, we can reverse the entire array.
- Reverse the first k elements and the remaining elements separately: After reversing the entire array, we need to reverse the first k elements and the remaining elements separately to get the rotated array.

**Code Explanation**

Let's take a look at the code to understand how we are implementing the above three steps.

function rotate(nums, k) {

const n = nums.length;

k = k % n;

reverse(nums, 0, n - 1);

reverse(nums, 0, k - 1);

reverse(nums, k, n - 1);

}

The **rotate** function takes** **two arguments: **nums **and** k**. The nums argument is the integer array that needs to be rotated, and the k argument is the non-negative integer that represents the number of steps to rotate the array.

In the first line of the rotate function, we get the length of the nums array and calculate the effective rotation index k by taking the modulus of k with the length of the array.

Next, we call the reverse function three times to get the rotated array:

function reverse(nums, start, end) {

while (start < end) {

const temp = nums[start];

nums[start] = nums[end];

nums[end] = temp;

start++;

end--;

}

}

The reverse function takes three arguments: nums, start, and end. The nums argument is the integer array that needs to be reversed. The `start` argument is the starting index of the array, and the end argument is the ending index of the array.

Inside the reverse function, we swap the values of nums[start] and nums[end] using a temporary variable temp. We continue swapping until the start index becomes greater than or equal to the end index.

**Complete Code:**

function rotate(nums, k) {

const n = nums.length;

k = k % n;

reverse(nums, 0, n - 1);

reverse(nums, 0, k - 1);

reverse(nums, k, n - 1);

function reverse(nums, start, end) {

while (start < end) {

const temp = nums[start];

nums[start] = nums[end];

nums[end] = temp;

start++;

end--;

}

return nums;

}

}

**Conclusion**

In this article, we have discussed an efficient approach to rotate an array to the right by k steps using JavaScript. This approach has a time **complexity of O(n)** and a **space complexity of O(1)**, where n is the length of the array. We have also explained the code step-by-step to help you understand the implementation.