Rotate Array LeetCode Solution

Problem – Rotate Array

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?

Rotate Array LeetCode Solution in Java

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

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

Rotate Array LeetCode Solution in C++

    class Solution 
    {
    public:
        void rotate(int nums[], int n, int k) 
        {
            if ((n == 0) || (k <= 0))
            {
                return;
            }
            
            // Make a copy of nums
            vector<int> numsCopy(n);
            for (int i = 0; i < n; i++)
            {
                numsCopy[i] = nums[i];
            }
            
            // Rotate the elements.
            for (int i = 0; i < n; i++)
            {
                nums[(i + k)%n] = numsCopy[i];
            }
        }
    };

Rotate Array LeetCode Solution in Python

class Solution:
    # @param nums, a list of integer
    # @param k, num of steps
    # @return nothing, please modify the nums list in-place.
    def rotate(self, nums, k):
        n = len(nums)
        k = k % n
        nums[:] = nums[n-k:] + nums[:n-k]
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