LeetCode-in-All

189. Rotate Array

Medium

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

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3

Output: [5,6,7,1,2,3,4]

Explanation:

rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4] 

Example 2:

Input: nums = [-1,-100,3,99], k = 2

Output: [3,99,-1,-100]

Explanation:

rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100] 

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

To solve the problem and rotate the array to the right by k steps, we can use multiple approaches. One efficient way is to use array reversal. This approach requires O(1) extra space.

Steps:

1. Reverse the Entire Array:
   • Reverse the entire array.
2. Reverse the First k Elements:
   • Reverse the first k elements of the array.
3. Reverse the Remaining Elements:
   • Reverse the remaining elements of the array after the first k elements.

Implementation:

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        def reverse(arr, start, end):
            while start < end:
                arr[start], arr[end] = arr[end], arr[start]
                start += 1
                end -= 1

        n = len(nums)
        k %= n

        # Reverse the entire array
        reverse(nums, 0, n - 1)

        # Reverse the first k elements
        reverse(nums, 0, k - 1)

        # Reverse the remaining elements
        reverse(nums, k, n - 1)

Explanation:

1. Reverse the Entire Array:
   • The reverse function is defined to reverse a portion of the array between indices start and end.
   • We call reverse(nums, 0, n - 1) to reverse the entire array.
2. Reverse the First k Elements:
   • We calculate k as k %= n to handle cases where k is greater than the length of the array.
   • We call reverse(nums, 0, k - 1) to reverse the first k elements of the array.
3. Reverse the Remaining Elements:
   • We call reverse(nums, k, n - 1) to reverse the remaining elements of the array after the first k elements.

By following these steps, the array will be rotated to the right by k steps. This approach ensures that the rotation is done in-place with O(1) extra space and is efficient with a time complexity of O(n).

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