Rotate Array Leetcode Problem 189 [Python Solution]

Are you ready to tackle another exciting LeetCode problem?

In this guide, we'll explore the Rotate Array problem (LeetCode Problem 189).

We'll break down the problem, discuss its constraints, explore the brute-force approach with Python code, and then delve into a more efficient solution that optimizes both time and space complexity.

By the end of this guide, you'll have a solid understanding of how to solve this problem and rotate an array efficiently.

Let's dive in!

Problem Overview

Question: Given an integer array nums, we need to rotate the array to the right by k steps, where k is a non-negative integer.

Let's illustrate this with an example:

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]

Explanation:
– Rotate 1 step 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]

Another example to illustrate:

Example 2:

Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]

Explanation:
– Rotate 1 step to the right: [99,-1,-100,3]
– Rotate 2 steps to the right: [3,99,-1,-100]

The problem seems clear now.

We are given an array, and we need to shift its elements to the right by k positions.

Understanding Constraints

Before we proceed, let's understand the constraints of this problem:

• 1 <= nums.length <= 10^5: The length of the input array nums will be between 1 and 100,000.
• -2^31 <= nums[i] <= 2^31 - 1: Each element in the input array nums is an integer within the range of -2^31 to 2^31 – 1.
• 0 <= k <= 10^5: The value of k is a non-negative integer.

Rotate Array LeetCode Problem Solution

1. Bruteforce Approach

A straightforward approach to solving this problem is to use extra space to create a new array and copy elements to the new array based on the new positions.

Here's a Python code solution that follows this approach:

def rotate(self, nums: List[int], k: int) -> None:
    """
    Do not return anything, modify nums in-place instead.
    """
    k = k % len(nums)  # Ensure k is within bounds.

result = [0] * len(nums)
    for i in range(len(nums)):
        new_index = (i + k) % len(nums)
        result[new_index] = nums[i]
    nums[:] = result

While this approach works, it uses extra memory, and its time complexity is O(n), where n is the length of the input array nums.

Let's explore a more efficient solution that solves the problem in O(1) space complexity and O(n) time complexity.

2. An Efficient Approach with Array Reversal

To achieve a more efficient solution, we'll use an approach that requires no extra memory and has a time complexity of O(n).

This efficient solution leverages the power of array reversal.

Here's the Python code:

def rotate(self, nums: List[int], k: int) -> None:
    """
    Do not return anything, modify nums in-place instead.
    """
    k = k % len(nums)  # Ensure k is within bounds.

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

    reverse(nums, 0, len(nums) - 1)       # Reverse the entire array.

reverse(nums, 0, k - 1)               # Reverse the first k elements.

reverse(nums, k, len(nums) - 1)       # Reverse the remaining elements.

Let's break down this efficient solution:

• First, we ensure that k is within bounds by taking k % len(nums).

• We define a helper function reverse(nums, start, end) that reverses the elements in the given range.

This helper function performs the array reversal in place.

• We perform three reversals:
  1. Reverse the entire array.
  2. Reverse the first k elements.
  3. Reverse the remaining elements.

By using this approach, we achieve the desired rotation of the array without using extra memory.

The time complexity of this solution is O(n), where n is the length of the input array nums.

Time and Space Complexity

Time Complexity:
– The brute-force solution has a time complexity of O(n), where n is the length of the input array nums.

• The efficient solution, which uses array reversal, also has a time complexity of O(n) since it performs three separate array reversals, each of which takes O(n) time.

Space Complexity:
– The brute-force solution uses extra memory to create a new array, so its space complexity is O(n).

• The efficient solution, which modifies the input array in place, uses no extra memory.

Hence, its space complexity is O(1).

Reasoning Behind Our Approach

In the efficient solution, we first perform a total reversal of the input array.

This is a key step as it effectively moves the last k elements to the beginning and the remaining elements to the end.

We then perform two more reversals: one for the first k elements and another for the remaining elements.

This ensures that the array is rotated to the right by k steps.

The reason this approach is efficient is that it leverages the power of array reversal, which can be done in place without requiring additional memory.

By carefully reversing different segments of the array, we achieve the desired rotation.

The modulus operation ensures that k remains within bounds, preventing unnecessary rotations.

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Conclusion

In this guide, we tackled the Rotate Array problem (LeetCode Problem 189).

We explored both a brute-force solution that uses extra memory and an efficient solution that utilizes array reversal for an optimal time and space complexity.

Remember, it's essential to understand the problem constraints, especially when working with large datasets.

We also discussed the reasoning behind our efficient solution, emphasizing the use of array reversal.

Now that you have a clear understanding of how to solve this problem, go ahead and practice implementing these solutions.

Feel free to comment, ask questions, make suggestions, and share this content with others who might find it helpful.

If you're ready to test your solution, you can find the problem on LeetCode here: Rotate Array on LeetCode.

Happy coding, and keep exploring the exciting world of algorithmic problem

-solving!