Rotate Array

Problem Description

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

Examples

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

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

Solution for Candy Distribution Problem

Intuition And Approach

To rotate an array to the right by k steps, we need to move the last k elements to the front and shift the rest of the elements to the right. A straightforward way to achieve this in-place (without using extra space for another array) is to use the reversal method.

Here is the step-by-step approach:

Adjust k:

If k is greater than the length of the array, rotating by k steps is the same as rotating by k % n steps (where n is the length of the array). This is because rotating by the length of the array brings it back to the original position. Calculate k = k % n.

Reverse the entire array:

By reversing the entire array, the last k elements (which we want to move to the front) will be at the beginning, but in reverse order. For example, reversing [1, 2, 3, 4, 5, 6, 7] gives [7, 6, 5, 4, 3, 2, 1].

Reverse the first k elements:

The first k elements are now the elements that were originally at the end of the array. Reverse these to restore their original order. Continuing the example, reversing the first 3 elements of [7, 6, 5, 4, 3, 2, 1] gives [5, 6, 7, 4, 3, 2, 1].

Reverse the remaining n - k elements:

Finally, reverse the rest of the array (from the k-th element to the end) to restore their order. In the example, reversing the elements from index 3 to 6 of [5, 6, 7, 4, 3, 2, 1] gives [5, 6, 7, 1, 2, 3, 4].

Code in Different Languages

Java

Written by @mahek0620

 
 class Solution {
 public void rotate(int[] nums, int k) {
     // Ensure k is within the bounds of the array length
     k = k % nums.length;
     
     // Reverse the entire array
     reverse(nums, 0, nums.length - 1);
     
     // Reverse the first k elements
     reverse(nums, 0, k - 1);
     
     // Reverse the remaining elements
     reverse(nums, k, nums.length - 1);
 }
 
 // Helper function to reverse elements in the array from start to end
 private void reverse(int[] nums, int start, int end) {
     while (start < end) {
         int temp = nums[start];
         nums[start] = nums[end];
         nums[end] = temp;
         start++;
         end--;
     }
 }

Python

Written by @mahek0620

 class Solution(object):
 def rotate(self, nums, k):
     k = 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, start, end):
     while start < end:
         nums[start], nums[end] = nums[end], nums[start]
         start += 1
         end -= 1

C++

Written by @mahek0620

 
  #include <vector>
  using namespace std;

  class Solution {
  public:
  void rotate(vector<int>& nums, int k) {
      // Ensure k is within the bounds of the array length
      k = k % nums.size();
      
      // Reverse the entire array
      reverse(nums, 0, nums.size() - 1);
      
      // Reverse the first k elements
      reverse(nums, 0, k - 1);
      
      // Reverse the remaining elements
      reverse(nums, k, nums.size() - 1);
  }
  
  private:
  void reverse(vector<int>& nums, int start, int end) {
      while (start < end) {
          int temp = nums[start];
          nums[start] = nums[end];
          nums[end] = temp;
          start++;
          end--;
      }
  }
};

References

• LeetCode Problem: House robber Problem
• Solution Link: House Robber Solution on LeetCode
• Authors GeeksforGeeks Profile: Mahek Patel