Rotate Array - Array - Medium - LeetCode
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Rotate Array - Array - Medium - LeetCode

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  • 1The problem involves rotating an array to the right by k steps, where k is a non-negative integer.
  • 2Three different methods can be used to solve the rotation problem, including an in-place solution with O(1) extra space.
  • 3The time complexity of the optimal solution is O(n), while the space complexity is O(1).

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Key Insight
AskGif

"The problem involves rotating an array to the right by k steps, where k is a non-negative integer."

Rotate Array - Array - Medium - LeetCode

Given an array, rotate the array to the right by k steps, where k is non-negative.

Follow up:

Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem. Could you do it in-place with O(1) extra space?

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 <= 2 * 104 -231 <= nums[i] <= 231 - 1 0 <= k <= 105

public class Solution {
 public void Rotate(int[] nums, int k) {
 k = k % nums.Length; 
 Reverse(ref nums,0, nums.Length -1 - k);
 Reverse(ref nums, nums.Length-k, nums.Length-1);
 Reverse(ref nums, 0, nums.Length-1);
 }
 
 private void Reverse(ref int[] nums, int start, int end){
 for(int i=start, j=end;i<j;i++,j--){
 Swap(ref nums,i,j);
 } 
 }
 
 private void Swap(ref int[] nums, int source, int destination){
 int temp = nums[source];
 nums[source] = nums[destination];
 nums[destination] = temp;
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

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sumitc91

Published on 18 November 2020 · 1 min read · 201 words

Part of AskGif Blog · coding

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