189. Rotate Array

Updated: 2024-03-12
2 min read
[Algorithms Medium]

LeetCode problem

Problem Statement

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

Naive Solution

A simple, but inefficient, approach would be to rotate the array k times. In each rotation, we shift every element of the array to the right by one and move the last element to the start of the array. This solution has a time complexity of O(n*k), where n is the number of elements in the array and k is the number of rotations. This is not an optimal solution, especially when we have a large k or a large array.


An efficient solution can be found by using array reversal. Here’s the plan:

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

This method allows us to achieve the desired output in O(n) time and O(1) space complexity.


Let’s break down the steps using an example: nums = [1,2,3,4,5,6,7], k = 3.

  1. Reverse the entire array: nums = [7,6,5,4,3,2,1].
  2. Reverse the first k elements: nums = [5,6,7,4,3,2,1].
  3. Reverse the remaining n-k elements: nums = [5,6,7,1,2,3,4].

As you can see, we get the expected output [5,6,7,1,2,3,4].


Here is the Python code that implements the aforementioned logic:

class Solution:
    def rotate(self, nums, k):

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

        n = len(nums)
        k = k % n  # in case k > len(nums)

        reverse(nums, 0, n - 1)  # 1
        reverse(nums, 0, k - 1)  # 2
        reverse(nums, k, n - 1)  # 3