Rotate Array LeetCode Solution | Easy Approach

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Rotate Array Given an array, 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

 Rotate Array Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  void rotate(vector<int>& nums, int k) {
    k %= nums.size();
    reverse(nums, 0, nums.size() - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.size() - 1);
  }

 private:
  void reverse(vector<int>& nums, int l, int r) {
    while (l < r)
      swap(nums[l++], nums[r--]);
  }
};

Java

 
 class Solution {
  public void rotate(int[] nums, int k) {
    k %= nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
  }

  private void reverse(int[] nums, int l, int r) {
    while (l < r)
      swap(nums, l++, r--);
  }

  private void swap(int[] nums, int l, int r) {
    final int temp = nums[l];
    nums[l] = nums[r];
    nums[r] = temp;
  }
}

Python

 class Solution:
  def rotate(self, nums: List[int], k: int) -> None:
    k %= len(nums)
    self.reverse(nums, 0, len(nums) - 1)
    self.reverse(nums, 0, k - 1)
    self.reverse(nums, k, len(nums) - 1)

  def reverse(self, nums: List[int], l: int, r: int) -> None:
    while l < r:
      nums[l], nums[r] = nums[r], nums[l]
      l += 1
      r -= 1

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