189. Rotate Array

Medium

Problem Description

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

Examples

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]
Jump to Solution: Python Java C++ JavaScript C#

Python Solution

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        n = len(nums)
        k = k % n  # Handle k > n case
        
        def reverse(start: int, end: int) -> None:
            while start < end:
                nums[start], nums[end] = nums[end], nums[start]
                start += 1
                end -= 1
        
        # Reverse entire array
        reverse(0, n - 1)
        # Reverse first k elements
        reverse(0, k - 1)
        # Reverse remaining elements
        reverse(k, n - 1)

Java Solution

class Solution {
    public void rotate(int[] nums, int k) {
        int n = nums.length;
        k = k % n;  // Handle k > n case
        
        reverse(nums, 0, n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);
    }
    
    private void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}

C++ Solution

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

JavaScript Solution

/**
 * @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 = k % n;  // Handle k > n case
    
    const reverse = (start, end) => {
        while (start < end) {
            [nums[start], nums[end]] = [nums[end], nums[start]];
            start++;
            end--;
        }
    };
    
    reverse(0, n - 1);
    reverse(0, k - 1);
    reverse(k, n - 1);
};

C# Solution

public class Solution {
    public void Rotate(int[] nums, int k) {
        int n = nums.Length;
        k = k % n;  // Handle k > n case
        
        Reverse(nums, 0, n - 1);
        Reverse(nums, 0, k - 1);
        Reverse(nums, k, n - 1);
    }
    
    private void Reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}

Complexity Analysis

Solution Explanation

This solution uses the reversal algorithm:

  1. First, we handle the case where k > n by using k = k % n
  2. Then we perform three reversals:
    1. Reverse the entire array
    2. Reverse the first k elements
    3. Reverse the remaining elements

Why this works: