The given code is a C++ implementation of a solution...

The given code is a C++ implementation of a solution to rotate an array to the right by k steps. Specifically, the function modifies the input array nums in-place. Here's a detailed breakdown of what the code does:

Function:

void rotate(vector<int>& nums, int k)

Input:

• nums: A vector (dynamic array) of integers that needs to be rotated.
• k: The number of steps by which to rotate the array to the right.

Step-by-step Explanation:

1. Reverse the Entire Array:

   int i=0;
   int j=nums.size()-1;
   while(i<j){
       swap(nums[i],nums[j]);
       i++;
       j--;
   }
   

   • The first while loop reverses the order of all elements in array nums.
   • For example, if nums = [1, 2, 3, 4, 5], after this loop, nums becomes [5, 4, 3, 2, 1].

2. Reverse the First k Elements:

   int s=0;
   int e=k-1;
   while(s<=e){
       swap(nums[s],nums[e]);
       s++;
       e--;
   }
   

   • The second while loop reverses the first k elements of the reversed array (the first k positions).
   • For example, if the reversed array is [5, 4, 3, 2, 1] and k = 2, this loop will reverse the first 2 elements: [4, 5, 3, 2, 1].

3. Reverse the Remaining Elements:

   int p=k;
   int t=nums.size()-1;
   while(p<=t){
       swap(nums[p],nums[t]);
       p++;
       t--;
   }
   

   • The third while loop reverses the remaining elements (from index k to the end of the array).
   • For example, continuing with the previous array [4, 5, 3, 2, 1] with k = 2, this loop will reverse the last nums.size() - k elements: [4, 5, 1, 2, 3].

4. Output: After all three steps, the array nums is rotated to the right by k steps.

Key Points:

• To ensure that the rotation is valid for cases where k is larger than the size of the array (nums.size()), this code implicitly assumes that k will be reduced modulo nums.size() elsewhere in the program (before this function is called). If not handled externally, you may add this line at the beginning of the function:

  k = k % nums.size();
  

• The function rotates the array in O(n) time complexity and uses O(1) extra space since it performs the operation in-place.

Example:

Input:

nums = [1, 2, 3, 4, 5, 6, 7];
k = 3;

Execution:

1. Reverse the entire array: nums = [7, 6, 5, 4, 3, 2, 1].
2. Reverse the first k elements: nums = [5, 6, 7, 4, 3, 2, 1].
3. Reverse the remaining elements: nums = [5, 6, 7, 1, 2, 3, 4].

Output:

nums = [5, 6, 7, 1, 2, 3, 4];

This is the rotated array after shifting it to the right by 3 steps.