Rotate Array

Question

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

Follow up: Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem. Could you do it in-place with O(1) extra space?

Example:

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]

Analyzation

Brute Force

Rotate the array k times, rotate 1 element each time.

The Code

public class Solution {
    public void rotate(int[] nums, int k) {
        int temp, previous;
        for (int i = 0; i < k; i++) {
            previous = nums[nums.length - 1];
            for (int j = 0; j < nums.length; j++) {
                temp = nums[j];
                nums[j] = previous;
                previous = temp;
            }
        }
    }
}

Time & Space Complexity

• Time Complexity: Each element has been rotated 1 step by k times so it is O(n * k).
• Space Complexity: Since we only use constant number of variables, the Space Complexity is O(1).

Optimization

Rotate 3 times

We know that k % n tail elements will be moved to the head. Therefore, rotate the whole array, then rotate the first k elements, finally rotate the last n - k elements.

class Solution {
    public void rotate(int[] nums, int k) {
        k %= nums.length;

        helper(nums, 0, nums.length - 1);
        helper(nums, 0, k - 1);
        helper(nums, k, nums.length - 1);
    }

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

Time & Space Complexity

• Time Complexity: Since we rotate n elements by 3 times so it is O(3n) which is O(n).
• Space Complexity: Since we only use constant number of variables, the Space Complexity is O(1).