# 189-rotate-array


Try it on <a href='https://leetcode.com/problems/189-rotate-array'>leetcode</a>

## Description
<div class="description">
<div><p>Given an array, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3
<strong>Output:</strong> [5,6,7,1,2,3,4]
<strong>Explanation:</strong>
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]
</pre>

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> nums = [-1,-100,3,99], k = 2
<strong>Output:</strong> [3,99,-1,-100]
<strong>Explanation:</strong> 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>Try to come up with as many solutions as you can. There are at least <strong>three</strong> different ways to solve this problem.</li>
	<li>Could you do it in-place with <code>O(1)</code> extra space?</li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        self.usinglistcomprehension(nums, k)

    def usinglistcomprehension(self, nums: List[int], k: int) -> None:
        k %= len(nums)
        nums[:] = nums[-k:] + nums[:-k]

    def usingreverse(self, nums: List[int], k: int) -> None:
        k %= len(nums)
        self.reverse(nums, 0, len(nums) - 1)
        self.reverse(nums, 0, k - 1)
        self.reverse(nums, k, len(nums) - 1)

    def usingextraArray(self, nums: List[int], k: int) -> None:
        n = len(nums)
        tmp = [0] * n
        for i in range(n):
            tmp[(i + k) % len(nums)] = nums[i]
        for i in range(n):
            nums[i] = tmp[i]

    def bruteforce(self, nums: List[int], k: int) -> None:
        for i in range(k):
            prev = nums[-1]
            for j in range(len(nums)):
                nums[j], prev = prev, nums[j]

    def reverse(self, nums, start, end):
        while start < end:
            nums[start], nums[end] = nums[end], nums[start]
            start += 1
            end -= 1

```