LeetCode Day-4

Hi there,

Welcome to Day 4 of my LeetCode journey! I'm thrilled to have you here as I embark on my coding journey. As I mentioned before, I'm dedicated to tackling one problem each day and documenting my progress in this blog.

Let's get started!

189. Rotate Array

Intuition :

integer array nums
rotate the array to the right by k

Approach :

Calculate the length of the array. Store it in n.
```
  n = len(nums)
```
If the length of the array is less than k, then there will be a full rotation. So, you can decrement k by n
```
  if n < k:
      k = k-n
```
If k is less than the length of the array, then, the last k elements of the array come first. This can be done by slicing. Then replace the numbers in num with the sliced array.
nums[n-k :] give the last k elements

nums[:n-k] gives the first k elements

  if (k < n) :
      temp = nums[n-k:] + nums[:n-k]
      for i in range(n):
          nums[i] = temp[i]

Else, you have to rotate the array to the right by one element, k number of times. This an be done using a while loop and a temporary array

  else :
      temp = nums.copy()
      while (k > 0):
          temp = [temp[-1]] + temp[:-1]
          k -= 1

      for i in range(n):
          nums[i] = temp[i]

Here is the full code :

  class Solution:
      def rotate(self, nums: List[int], k: int) -> None:
          """
          Do not return anything, modify nums in-place instead.
          """

          n = len(nums)

          if n < k:
              k = k-n

          if (k < n) :
              temp = nums[n-k:] + nums[:n-k]
              for i in range(n):
                  nums[i] = temp[i]

          else :
              temp = nums.copy()
              while (k > 0):
                  temp = [temp[-1]] + temp[:-1]
                  k -= 1

              for i in range(n):
                  nums[i] = temp[i]

Time Complexity : O(n)

And that concludes Day 4 of my LeetCode Journey. If you have any ideas for enhancing my progress, don't hesitate to share them. Your encouragement and involvement means a lot to me.

Happy Coding!