# 189. Rotate Array

## Description

Given an integer array nums, 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

• 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?

## Solutions

Solution 1: Reverse three times

We can assume the length of the array is $n$ and calculate the actual number of steps needed by taking the module of $k$ and $n$, which is $k \bmod n$.

Next, let us reverse three times to get the final result:

1. Reverse the entire array.
2. Reverse the first $k$ elements.
3. Reverse the last $n - k$ elements.

For example, for the array $[1, 2, 3, 4, 5, 6, 7]$, $k = 3$, $n = 7$, $k \bmod n = 3$.

1. In the first reverse, reverse the entire array. We get $[7, 6, 5, 4, 3, 2, 1]$.
2. In the second reverse, reverse the first $k$ elements. We get $[5, 6, 7, 4, 3, 2, 1]$.
3. In the third reverse, reverse the last $n - k$ elements. We get $[5, 6, 7, 1, 2, 3, 4]$, which is the final result.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
private int[] nums;

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

private void reverse(int i, int j) {
for (; i < j; ++i, --j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
}

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

• class Solution:
def rotate(self, nums: List[int], k: int) -> None:
k %= len(nums)
nums[:] = nums[-k:] + nums[:-k]

############

class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
n = len(nums)
k %= n
if n < 2 or k == 0:
return
nums[:] = nums[::-1]
nums[:k] = nums[:k][::-1]
nums[k:] = nums[k:][::-1]

############

class Solution(object):
def rotate(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: void Do not return anything, modify nums in-place instead.
"""
if len(nums) == 0 or k == 0:
return

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

n = len(nums) - 1
k = k % len(nums)
reverse(0, n - k, nums)
reverse(n - k + 1, n, nums)
reverse(0, n, nums)


• func rotate(nums []int, k int) {
n := len(nums)
k %= n
reverse := func(i, j int) {
for ; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
}
reverse(0, n-1)
reverse(0, k-1)
reverse(k, n-1)
}

• /**
Do not return anything, modify nums in-place instead.
*/
function rotate(nums: number[], k: number): void {
const n: number = nums.length;
k %= n;
const reverse = (i: number, j: number): void => {
for (; i < j; ++i, --j) {
const t: number = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
};
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
}


• /**
* @param {number[]} nums
* @param {number} k
* @return {void} Do not return anything, modify nums in-place instead.
*/
var rotate = function (nums, k) {
const n = nums.length;
k %= n;
const reverse = (i, j) => {
for (; i < j; ++i, --j) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
};
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
};


• public class Solution {
private int[] nums;

public void Rotate(int[] nums, int k) {
this.nums = nums;
int n = nums.Length;
k %= n;
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
}

private void reverse(int i, int j) {
for (; i < j; ++i, --j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
}

• impl Solution {
pub fn rotate(nums: &mut Vec<i32>, k: i32) {
let n = nums.len();
let k = (k as usize) % n;
nums.reverse();
nums[..k].reverse();
nums[k..].reverse();
}
}