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# Rotate Array

## Description

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

**Example 1:**

```
Input: [1,2,3,4,5,6,7] and 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: [-1,-100,3,99] and 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]
```

**Note:**

* 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?

## Solutions

### My solution (also optimal)

Idea: directly put numbers in their new positions.

```cpp
class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        if(n == 0 || k % n == 0) return;
        k %= n;
        int count = 0; // number of elements in correct position
        int start = 0;
        while(count < n){
            int i = start;
            int prev = nums[i]; // value of number in previous index
            do{
                i = (i + k) % n; // destination index
                swap(prev, nums[i]);
                ++count;
            }while(i != start);
            ++start;
        }
    }
};
```

### Easier to understand solution

Idea: the objective is to move the first `n - k` numbers to the last `n - k` positions, and move the last `k` numbers to the first `k` positions. If we reverse `nums`, then the original first `n - k` numbers and the last `k` numbers are in correct ranges. Then we reverse the two ranges to make all numbers in correct positions.

**E.g.**,

Original array: `[1, 2, 3, 4, 5, 6]` , k = 4

Objective: `[3, 4, 5, 6, 1, 2]`

After first reversing: `[6, 5, 4, 3, 2, 1]`.`[6, 5, 4, 3]` and `[2, 1]` are in correct ranges but in reversed order.

```cpp
class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        if(n == 0 || k % n == 0) return;
        k %= n;
        reverse(nums.begin(), nums.end());
        reverse(nums.begin(), nums.begin() + k);
        reverse(nums.begin() + k, nums.end());
    }
};
```