Given an integer array nums, 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 - 10 <= 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?
Step 1: n is assigned the length of the input array nums. This determines the number of elements in the array.
Step 2: k is taken modulo n to ensure that it is within the range (0, n). This is because if k is larger than n, rotating the array by k positions is equivalent to rotating it by k % n positions.
Step 3: If k is negative, it is adjusted by adding n to it. This is to handle cases where k is a negative number. Adding n to a negative k effectively rotates the array to the left.
Step 4: An auxiliary array temp of the same length as nums is created. This array will be used to store the rotated elements.
Step 5: The loop copies the elements from nums to temp with a right shift of k positions. The modulo operation is used to ensure that the indices wrap around if they exceed the length of the array.
Step 6: Finally, another loop copies the rotated elements from the temp array back to the original nums array.
The code performs two separate loops through the array, each with n iterations. Therefore, the time complexity is O(n).
The code uses an additional temp array of the same size as the input array nums. So the space complexity is O(n) as well.