0189-rotate-array

Try it on leetcode

Description

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]
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]

 

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

 

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Solution(Python)

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

        
    # poping each element and appending in front
    # Time=complexity: O(n,k)
    # Space complexity: O(n)
    def bruteforce(self, nums, k):
        k %= len(nums)
        for _ in range(k):
            nums.insert(0, nums.pop())

    # use list com,phgrehension inspace
    # Time=complexity: O(n)
    # Space complexity: O(n)
    def listcomprehension(self, nums, k):
        k = k% len(nums)
        nums[:] = nums[-k:] + nums[:-k]


    # use inplace
    # Time=complexity: O(n)
    # Space complexity: O(1)
    def inplaceoptimal(self, nums, k):
        n = len(nums)
        k %= n
        def reverse(l ,r):
            while l < r:
                nums[l], nums[r] = nums[r], nums[l]
                l+=1
                r -= 1
        reverse(0,n -1 )
        reverse(0, k -1)
        reverse(k , n-1)