Rotate Array
In the world of coding and algorithmic challenges, the task of rotating an array might seem straightforward at first glance. However, as we delve deeper, we find that efficiency and optimization play a crucial role in crafting the ideal solution. Today, we will explore two different approaches to solve the “Rotate Array” problem, where we are given an integer array nums and the task is to rotate the array to the right by k steps.
Problem Statement
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]
Example 2:
- Input: nums = [-1,-100,3,99], k = 2
- Output: [3,99,-1,-100]
Approach 1: Brute Force
First solution to this problem takes a brute-force approach:
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
for i in range(k):
nums.insert(0, nums[-1])
nums.pop(-1)
In this approach, we rotate the array one step at a time, k times. Each rotation involves moving the last element of the array to the front, effectively rotating the array by one position. While this method is straightforward and easy to understand, its efficiency is less than ideal. The runtime for this approach is 2240 ms, which is quite slow. The reason for this inefficiency is the repeated shifting of elements and the use of insert() and pop() methods in each iteration.
Approach 2: Efficient Rotation
To optimize our solution, we need to reduce the number of operations performed on the array.
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
if len(nums) == 0:
return []
if k == 0:
return nums
if len(nums) <= k:
for i in range(k):
nums.insert(0, nums.pop())
return
nums1 = nums[:len(nums) - k]
nums2 = nums[len(nums) - k:]
num = nums1[::-1] + nums2[::-1]
nums[:] = num[::-1]
In this approach, we first handle edge cases like empty arrays or no rotation needed (k == 0). If k is greater than the length of the array, we simply rotate the entire array k times. The key optimization happens when we split the array into two parts: the part that moves to the front (nums2) and the remaining part (nums1). We then reverse each part separately and concatenate them. Finally, we reverse the entire concatenated array to achieve the rotated array. This method significantly reduces the number of operations, resulting in a much faster runtime of 150 ms.
Why is the Second Approach Better?
The second approach is superior for a few reasons:
- Fewer Operations: It minimizes the number of shifting operations required, which is the primary source of inefficiency in the first approach.
- Better Use of Slicing and Reversing: Python’s slicing and list reversal are optimized and faster than manually moving elements one by one.
- Handles Edge Cases More Efficiently: By considering cases like empty arrays or rotations greater than the array’s length, the second approach is more robust
When solving array manipulation problems like this, it’s essential to think beyond the brute-force method. Optimizing your solution can lead to significant improvements in runtime and efficiency, as we’ve seen in our exploration of the “Rotate Array” problem. Remember, the key to optimization often lies in minimizing operations and making smart use of the language’s built-in functionalities.
