leetcode-189-rotate-array.swift

/**

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https://leetcode.com/problems/rotate-array/


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 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 <= 2 * 10^4
It's guaranteed that nums[i] fits in a 32 bit-signed integer.
k >= 0

*/

/**
Big O Annotation
Time complexity O(kn) k for the iterations and n for the insert at 0 operation.
Space complexity O(1) where not using any extra space but the popped element.
*/
func rotate(_ nums: inout [Int], _ k: Int) {
        
    /**
    Without elements in nums 
    there's nothing to do...
    */
    guard nums.count > 0 else {
        return 
    }
    
    // Iterate through 0 until k-1
    for _ in 0..<k {

        // Pop the last element
        let last = nums.popLast()!

        // And insert at index 0
        nums.insert(last, at: 0)
    }
}